Important Note

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Exercise 7.1

Find an antiderivative (or integral) of the following functions by the method of inspection.

1. sin 2x

Solun:- Let y = sin 2x

We know that

{"font":{"color":"#000000","size":12,"family":"Arial"},"code":"$\\diff{\\left(\\cos2x\\right)}{x}=-2\\sin2x$","id":"2-0","type":"$","ts":1599821005525,"cs":"F/eg5C58QRPw8OmAvqi0+w==","size":{"width":172,"height":28}}

{"font":{"family":"Arial","size":11,"color":"#000000"},"code":"\\begin{align*}\n{\\frac{-1}{2}\\diff{\\left(\\cos2x\\right)}{x}}&={\\sin2x}\\\\\n{\\diff{\\left(\\frac{-1}{2}\\cos2x\\right)}{x}}&={\\sin2x}\t\n\\end{align*}","id":"3-0","type":"align*","ts":1599820982937,"cs":"YlSJDpdfXE+GVOO46GRezA==","size":{"width":178,"height":86}}

Hence the Antiderivative of sin2x is -1/2cos2x.

2. cos 3x

Solun:- Let y = cos 3x

We know that

{"font":{"family":"Arial","color":"#000000","size":12},"code":"$\\diff{\\left(\\sin 3x\\right)}{x}=3\\cos 3x$","id":"2-1-0","type":"$","ts":1599821431019,"cs":"bM2sNNuLZ0iwokBDx85UvQ==","size":{"width":156,"height":28}}

{"font":{"color":"#000000","size":11,"family":"Arial"},"code":"\\begin{align*}\n{\\frac{1}{3}\\diff{\\left(\\sin3x\\right)}{x}}&={\\cos3x}\\\\\n{\\diff{\\left(\\frac{1}{3}\\sin3x\\right)}{x}}&={\\cos3x}\t\n\\end{align*}","id":"3-1-0","type":"align*","ts":1599822208126,"cs":"rOhNgKjsV3QzdNG+0Y0W2g==","size":{"width":164,"height":88}}

Hence the Antiderivative of cos3x is 1/3sin3x.

3. e2x

Solun:- Let y = e2x

We know that

{"font":{"color":"#000000","size":12,"family":"Arial"},"id":"2-1-1-0","code":"$\\diff{\\left(e^{2x}\\right)}{x}=2e^{2x}$","type":"$","ts":1599822673571,"cs":"/C/QFQXaaSWvDrI1/iZgrQ==","size":{"width":112,"height":30}}

{"font":{"family":"Arial","size":11,"color":"#000000"},"id":"3-1-1-0-0","type":"align*","code":"\\begin{align*}\n{\\frac{1}{2}\\diff{\\left(e^{2x}\\right)}{x}}&={e^{2x}}\\\\\n{\\diff{\\left(\\frac{1}{2}e^{2x}\\right)}{x}}&={e^{2x}}\t\n\\end{align*}","ts":1599822752676,"cs":"N+sX770p6vLVjy8BGY6jwA==","size":{"width":121,"height":90}}

Hence the Anti-Derivative of e2x is 1/2e2x.

4. (ax + b)2

Solun:- Let y = (ax + b)2

We know that

{"font":{"family":"Arial","color":"#000000","size":12},"id":"2-1-1-1","code":"$\\diff{\\left[\\left(ax+b\\right)^{3}\\right]}{x}=3\\left(ax+b\\right)^{2}\\times a$","type":"$","ts":1599986696937,"cs":"e5Uu/b0fDpqEdiPukl8Crw==","size":{"width":233,"height":40}}

{"code":"\\begin{align*}\n{\\frac{1}{3a}\\diff{\\left[\\left(ax+b\\right)^{3}\\right]}{x}}&={\\left(ax+b\\right)^{2}}\\\\\n{\\diff{\\left[\\frac{1}{3a}\\left(ax+b\\right)^{3}\\right]}{x}}&={\\left(ax+b\\right)^{2}}\t\n\\end{align*}","id":"3-1-1-1","font":{"color":"#000000","size":11,"family":"Arial"},"type":"align*","ts":1599986724726,"cs":"NtBqQ/CpKkaJ39LgW6v1sg==","size":{"width":228,"height":112}}

Hence the Antiderivative of (ax + b)2 is 1/3(ax + b)3.

5. sin2x - 4e3x

Solun:- Let y = e2x

We know that

{"font":{"size":11,"family":"Arial","color":"#000000"},"id":"4","code":"\\begin{align*}\n{\\diff{\\left(\\frac{-1}{2}\\cos2x-\\frac{4}{3}e^{3x}\\right)}{x}}&={\\frac{-1}{2}\\left(-2\\sin2x\\right)-\\frac{4}{3}\\left(3e^{3x}\\right)}\\\\\n{\\diff{\\left(\\frac{-1}{2}\\cos2x-\\frac{4}{3}e^{3x}\\right)}{x}}&={\\sin2x-4e^{3x}}\t\n\\end{align*}","type":"align*","ts":1599827355170,"cs":"Eb/cWpEcrtaC+M4a5mCt4g==","size":{"width":392,"height":92}}

Hence the Antiderivative of sin2x - 4e3x is -1/2cos2x - 4/3e3x .

Find the following integrals in Exercises 6 to 20:

{"type":"$","id":"5-0-0","font":{"size":12,"color":"#000000","family":"Arial"},"code":"$6.\\,\\int_{}^{}\\left(4e^{3x}+1\\right)dx$","ts":1599898437954,"cs":"oB5ronvFkPKchWhF0ZyG5g==","size":{"width":148,"height":24}}

Solun:- Let f(x) = {"type":"$","id":"5-1-0","font":{"color":"#000000","size":11,"family":"Arial"},"code":"$\\left(4e^{3x}+1\\right)$","ts":1599987343117,"cs":"vIA2TGx5yXjK0VHN0fsH7Q==","size":{"width":80,"height":21}}

{"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(4e^{3x}+1\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}4e^{3x}dx+\\int_{}^{}1.dx}\t\n\\end{align*}","id":"6-0","ts":1599898991419,"cs":"VPOxtDu2rzn74ItYec2icw==","size":{"width":220,"height":76}}

We know that 

{"code":"\\begin{align*}\n{\\int_{}^{}e^{x}dx}&={e^{x}+c}\\\\\n{\\int_{}^{}1.dx\\,\\,}&={x+c}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"id":"7-0","ts":1599899316299,"cs":"u4m0GmyBBCaNG6nR0Kph/A==","size":{"width":116,"height":76}}

{"code":"$\\int_{}^{}f\\left(x\\right)dx=\\frac{4e^{3x}}{3}+x+c$","id":"8-0","font":{"color":"#000000","size":11,"family":"Arial"},"type":"$","ts":1599899547054,"cs":"dnJBH0l6GVXFEdmngNUqbg==","size":{"width":186,"height":24}}

{"id":"5-0-1-0-0","font":{"color":"#000000","family":"Arial","size":12},"type":"$","code":"$7.\\,\\int_{}^{}x^{2}\\left(1-\\frac{1}{x^{2}}\\right)dx$","ts":1599899712933,"cs":"Ax+wtyPCMCXJ4Pfa+lECAA==","size":{"width":158,"height":24}}

Solun:- Let f(x) = {"id":"5-0-1-1","code":"$x^{2}\\left(1-\\frac{1}{x^{2}}\\right)$","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","ts":1599987368741,"cs":"rS63sWegAhyVrVU0+YMGPQ==","size":{"width":74,"height":20}}

{"font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","id":"6-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(x^{2}-1\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}x^{2}dx-\\int_{}^{}1.dx}\t\n\\end{align*}","ts":1599899947159,"cs":"Xo2sKR5RIVBXHXXVOIlavQ==","size":{"width":208,"height":76}}

We know that 

{"id":"7-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}x^{n}dx}&={\\frac{x^{n+1}}{n+1}+c}\\\\\n{\\int_{}^{}1.dx\\,\\,}&={x+c}\t\n\\end{align*}","ts":1599900032649,"cs":"8BaRrf5UbvlUwuZWZgUp6Q==","size":{"width":144,"height":78}}

{"id":"8-1-0","font":{"size":11,"color":"#000000","family":"Arial"},"code":"$\\int_{}^{}f\\left(x\\right)dx=\\frac{x^{3}}{3}-x+c$","type":"$","ts":1599900109050,"cs":"ih/6BoXQj8SQ18gAI/e4Aw==","size":{"width":176,"height":24}}

{"type":"$","font":{"color":"#000000","family":"Arial","size":12},"id":"5-0-1-0-1-0-0","code":"$8.\\,\\int_{}^{}\\left(ax^{2}+bx+c\\right)dx$","ts":1599900244954,"cs":"H83WeAaYdL9ZRO3ZhVg0SQ==","size":{"width":188,"height":24}}

Solun:- Let f(x) = {"type":"$","id":"5-0-1-0-1-1","code":"$\\left(ax^{2}+bx+c\\right)$","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1599987388869,"cs":"SFBQ/c3wzoZo9cy3XXOyEA==","size":{"width":100,"height":18}}

{"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(ax^{2}+bx+c\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}ax^{2}dx+\\int_{}^{}bxdx+\\int_{}^{}cdx}\t\n\\end{align*}","id":"6-1-1-0","ts":1599902948751,"cs":"yTb/daqieQmgKSFV0aTcsQ==","size":{"width":284,"height":76}}

We know that 

{"id":"7-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}x^{n}dx}&={\\frac{x^{n+1}}{n+1}+c}\\\\\n{\\int_{}^{}1.dx\\,\\,}&={x+c}\t\n\\end{align*}","ts":1599900032649,"cs":"7p23xGkFH4puM1EKaEJjnA==","size":{"width":144,"height":78}}

{"code":"$\\int_{}^{}f\\left(x\\right)dx=\\frac{ax^{3}}{3}+\\frac{bx^{2}}{2}+cx+C$","type":"$","id":"8-1-1-0","font":{"family":"Arial","size":11,"color":"#000000"},"ts":1599903052907,"cs":"elQnBW4nusCNgn5TtzLZWQ==","size":{"width":245,"height":24}}

{"type":"$","code":"$9.\\,\\int_{}^{}\\left(2x^{2}+e^{x}\\right)dx$","id":"5-0-1-0-1-0-1-0-0","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1599903169211,"cs":"JXNTpj5LuNk9iteJdgepBQ==","size":{"width":152,"height":24}}

Solun:- Let f(x) = {"font":{"size":10,"color":"#000000","family":"Arial"},"id":"5-0-1-0-1-0-1-1-0","code":"$\\left(2x^{2}+e^{x}\\right)$","type":"$","ts":1599987411308,"cs":"fRTydVyJ9um6Zcy7hkjnIw==","size":{"width":72,"height":18}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(2x^{2}+e^{x}\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}2x^{2}dx+\\int_{}^{}e^{x}dx}\t\n\\end{align*}","type":"align*","id":"6-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1599903279933,"cs":"J0xS9qDD3ddqp0gDAMolxA==","size":{"width":218,"height":76}}

We know that 

{"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}x^{n}dx}&={\\frac{x^{n+1}}{n+1}+c}\\\\\n{\\int_{}^{}e^{x}dx\\,\\,}&={e^{x}+c}\t\n\\end{align*}","id":"7-1-1-1-0","ts":1599903358053,"cs":"qHSo3ObffaC4k8ZeeABRxQ==","size":{"width":148,"height":78}}

{"id":"8-1-1-1-0","font":{"size":11,"color":"#000000","family":"Arial"},"type":"$","code":"$\\int_{}^{}f\\left(x\\right)dx=\\frac{2x^{3}}{3}+e^{x}+C$","ts":1599903385064,"cs":"F1OvljL7gGqoEikByNzrmw==","size":{"width":196,"height":24}}

{"font":{"family":"Arial","color":"#000000","size":12},"id":"5-0-1-0-1-0-1-0-1-0-0","type":"$","code":"$10.\\,\\int_{}^{}\\left({\\sqrt[]{x}}-\\frac{1}{{\\sqrt[]{x}}}\\right)^{2}dx$","ts":1599904181962,"cs":"b5gsvihLqp5TTRD2YcQ87g==","size":{"width":188,"height":40}}

Solun:- Let f(x) = {"font":{"size":10,"family":"Arial","color":"#000000"},"id":"5-0-1-0-1-0-1-0-1-1-0","type":"$","code":"$\\left({\\sqrt[]{x}}-\\frac{1}{{\\sqrt[]{x}}}\\right)^{2}$","ts":1599904252786,"cs":"/r/UA/PuvzalsrAImKvkYw==","size":{"width":89,"height":32}}

{"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left({\\sqrt[]{x}}-\\frac{1}{{\\sqrt[]{x}}}\\right)^{2}dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(x+\\frac{1}{x}-2\\right)dx}\t\n\\end{align*}","type":"align*","id":"6-1-1-1-1-0","ts":1599904424442,"cs":"3vCVw3DIZTPng8agjf64yw==","size":{"width":225,"height":84}}

We know that 

{"type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"id":"7-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}x^{n}dx}&={\\frac{x^{n+1}}{n+1}+c}\\\\\n{\\int_{}^{}\\frac{1}{x}dx\\,\\,}&={\\log_{}x+c}\t\n\\end{align*}","ts":1599904475331,"cs":"wuPi6WMYGtSgfqNtsO8VbA==","size":{"width":148,"height":78}}

{"type":"$","font":{"color":"#000000","size":11,"family":"Arial"},"code":"$\\int_{}^{}f\\left(x\\right)dx=\\frac{x^{2}}{2}+\\log_{}\\left|x\\right|-2x+C$","id":"8-1-1-1-1-0","ts":1599904559669,"cs":"NxJUi9pDrDVb2+3+GiJP1A==","size":{"width":258,"height":24}}

{"font":{"family":"Arial","color":"#000000","size":12},"code":"$11.\\,\\int_{}^{}\\frac{x^{3}+5x^{2}-4}{x^{2}}dx$","type":"$","id":"5-0-1-0-1-0-1-0-1-0-1-0","ts":1599904646239,"cs":"+wxBrj3fzI+jqi2oYRCV4g==","size":{"width":146,"height":28}}

Solun:- Let f(x) = {"id":"5-0-1-0-1-0-1-0-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"code":"$\\frac{x^{3}+5x^{2}-4}{x^{2}}$","type":"$","ts":1599904696472,"cs":"iWNjCLfzw+5RxKRvZMWtuw==","size":{"width":56,"height":21}}

{"font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","id":"6-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x^{3}+5x^{2}-4}{x^{2}}dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(x+5-\\frac{4}{x^{2}}\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(x+5-4x^{-2}\\right)dx}\t\n\\end{align*}","ts":1599904795421,"cs":"qdoKrMQz4wlNE3YyPZE11A==","size":{"width":224,"height":120}}

We know that 

{"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}x^{n}dx}&={\\frac{x^{n+1}}{n+1}+c}\\\\\n{\\int_{}^{}1dx\\,\\,}&={x+c}\t\n\\end{align*}","id":"7-1-1-1-1-1-0","ts":1599904863262,"cs":"FOb8A503dWUkQt+FL0gfgQ==","size":{"width":144,"height":78}}

{"type":"align*","font":{"family":"Arial","color":"#000000","size":11},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}}{2}+5x-4\\frac{x^{-1}}{-1}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{2}}{2}+5x+\\frac{4}{x}+C}\t\n\\end{align*}","id":"8-1-1-1-1-1-0","ts":1599904970716,"cs":"Q3AiShhuaoWzEo6CsbNVPg==","size":{"width":269,"height":92}}

{"id":"5-0-1-0-1-0-1-0-1-0-1-1-0","type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"code":"$12.\\,\\int_{}^{}\\frac{x^{3}+3x+4}{{\\sqrt[]{x}}}dx$","ts":1599905056219,"cs":"mGicBkiH42WhhLtCWS0WHw==","size":{"width":140,"height":32}}

Solun:- Let f(x) = {"code":"$\\frac{x^{3}+3x+4}{{\\sqrt[]{x}}}$","font":{"family":"Arial","color":"#000000","size":10},"id":"5-0-1-0-1-0-1-0-1-1-1-1-0","type":"$","ts":1599905092181,"cs":"qvgy+GZjigHHyZaLJo7/FQ==","size":{"width":52,"height":24}}

{"id":"6-1-1-1-1-1-1-0","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x^{3}+3x+4}{{\\sqrt[]{x}}}dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(x^{2}{\\sqrt[]{x}}+3{\\sqrt[]{x}}+\\frac{4}{{\\sqrt[]{x}}}\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(x^{\\frac{5}{2}}+3x^{\\frac{1}{2}}+4x^{\\frac{-1}{2}}\\right)dx}\t\n\\end{align*}","ts":1599905254821,"cs":"PA31LE0qgNECanls5h0h6A==","size":{"width":281,"height":124}}

We know that 

{"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}x^{n}dx}&={\\frac{x^{n+1}}{n+1}+c}\\\\\n{\\int_{}^{}1dx\\,\\,}&={x+c}\t\n\\end{align*}","id":"7-1-1-1-1-1-1-0","ts":1599904863262,"cs":"puj47RuOJPE8/SmxRGzKqg==","size":{"width":144,"height":78}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{\\frac{7}{2}}}{\\frac{7}{2}}+\\frac{3x^{\\frac{3}{2}}}{\\frac{3}{2}}+\\frac{4x^{\\frac{1}{2}}}{\\frac{1}{2}}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{2x^{\\frac{7}{2}}}{7}+\\frac{2\\times3x^{\\frac{3}{2}}}{3}+8x^{\\frac{1}{2}}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{2x^{\\frac{7}{2}}}{7}+2x^{\\frac{3}{2}}+8{\\sqrt[]{x}}+C}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"8-1-1-1-1-1-1-0","type":"align*","ts":1599905694554,"cs":"g10hAB+Ul4ZWZkkjvXpBgg==","size":{"width":278,"height":134}}

{"code":"$13.\\,\\int_{}^{}\\frac{x^{3}-x^{2}+x-1}{x-1}dx$","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"id":"5-0-1-0-1-0-1-0-1-0-1-1-1-0","ts":1599906045734,"cs":"vDs0TBSdzSKwbvxYfA250w==","size":{"width":158,"height":28}}

Solun:- Let f(x) = {"type":"$","id":"5-0-1-0-1-0-1-0-1-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\frac{x^{3}-x^{2}+x-1}{x-1}$","ts":1599906069447,"cs":"NYI04DXEJwDJXS6TjELLPw==","size":{"width":66,"height":21}}

{"id":"6-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x^{3}-x^{2}+x-1}{x-1}dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{x^{2}\\left(x-1\\right)+\\left(x-1\\right)}{x-1}dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\left(x-1\\right)\\left(x^{2}+1\\right)}{x-1}dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(x^{2}+1\\right)dx}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"ts":1599906279394,"cs":"7qZESFA7gzfJoSQttea4/w==","size":{"width":264,"height":166}}

We know that 

{"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}x^{n}dx}&={\\frac{x^{n+1}}{n+1}+c}\\\\\n{\\int_{}^{}1dx\\,\\,}&={x+c}\t\n\\end{align*}","id":"7-1-1-1-1-1-1-1-0","ts":1599904863262,"cs":"lKZHnkY1IElsWS3wHQ9Cgg==","size":{"width":144,"height":78}}

{"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"id":"8-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{3}}{3}+x+C}\t\n\\end{align*}","ts":1599906332059,"cs":"xVxwyCp/au/G7+rPHUid0g==","size":{"width":169,"height":37}}

{"code":"$14.\\,\\int_{}^{}\\left(1-x\\right){\\sqrt[]{x}}dx$","type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"id":"5-0-1-0-1-0-1-0-1-0-1-1-1-1-0-0","ts":1599906411126,"cs":"0DYWPCXZTgD0k+WKZrDuCw==","size":{"width":160,"height":24}}

Solun:- Let f(x) = {"code":"$\\left(1-x\\right){\\sqrt[]{x}}$","font":{"size":11,"color":"#000000","family":"Arial"},"id":"5-0-1-0-1-0-1-0-1-0-1-1-1-1-1-0","type":"$","ts":1599906445971,"cs":"JNckwmqYLBmj/FUPIa4ZtA==","size":{"width":80,"height":20}}

{"id":"6-1-1-1-1-1-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(1-x\\right){\\sqrt[]{x}}dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left({\\sqrt[]{x}}-x{\\sqrt[]{x}}\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(x^{\\frac{1}{2}}-x^{\\frac{3}{2}}\\right)dx}\t\n\\end{align*}","ts":1599906607625,"cs":"Mr47ZCdy15jw42AqSkcoAA==","size":{"width":209,"height":116}}

We know that 

{"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}x^{n}dx}&={\\frac{x^{n+1}}{n+1}+c}\\\\\n{\\int_{}^{}1dx\\,\\,}&={x+c}\t\n\\end{align*}","id":"7-1-1-1-1-1-1-1-1","ts":1599904863262,"cs":"OaX+PTbJOqSZLL+aNsmwSw==","size":{"width":144,"height":78}}

{"id":"8-1-1-1-1-1-1-1-1-0","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{\\frac{3}{2}}}{\\frac{3}{2}}-\\frac{x^{\\frac{5}{2}}}{\\frac{5}{2}}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{2x^{\\frac{3}{2}}}{3}-\\frac{2x^{\\frac{5}{2}}}{5}+C}\t\n\\end{align*}","ts":1599906743160,"cs":"Q12nr6FZGpdCfA9lvMKoCA==","size":{"width":205,"height":89}}

{"type":"$","code":"$15.\\,\\int_{}^{}{\\sqrt[]{x}}\\left(3x^{2}+2x+3\\right)dx$","font":{"family":"Arial","color":"#000000","size":12},"id":"5-0-1-0-1-0-1-0-1-0-1-1-1-1-0-1-0","ts":1599906916847,"cs":"lL8Yi6yyXExZvvtdqqHbbA==","size":{"width":232,"height":24}}

Solun:- Let f(x) = {"id":"5-0-1-0-1-0-1-0-1-0-1-1-1-1-1-1-0","font":{"size":11,"color":"#000000","family":"Arial"},"code":"${\\sqrt[]{x}}\\left(3x^{2}+2x+3\\right)$","type":"$","ts":1599907040341,"cs":"GKiTeARNNz1JcXMLo7GNYA==","size":{"width":141,"height":21}}

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"6-1-1-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}{\\sqrt[]{x}}\\left(3x^{2}+2x+3\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(3x^{\\frac{5}{2}}+2x^{\\frac{3}{2}}+3x^{\\frac{1}{2}}\\right)dx}\t\n\\end{align*}","ts":1599907143408,"cs":"7k/nuHUA0kN0Ze9d7oiuwQ==","size":{"width":260,"height":76}}

We know that 

{"font":{"family":"Arial","color":"#000000","size":10},"id":"7-1-1-1-1-1-1-1-2","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}x^{n}dx}&={\\frac{x^{n+1}}{n+1}+c}\t\n\\end{align*}","ts":1599907474905,"cs":"lsEDNgfI+OtCLgkN7q3beQ==","size":{"width":144,"height":37}}

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"8-1-1-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{3x^{\\frac{7}{2}}}{\\frac{7}{2}}+\\frac{2x^{\\frac{5}{2}}}{\\frac{5}{2}}+\\frac{3x^{\\frac{3}{2}}}{\\frac{3}{2}}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{6x^{\\frac{7}{2}}}{7}+\\frac{4x^{\\frac{5}{2}}}{5}+2x^{\\frac{3}{2}}+C}\t\n\\end{align*}","ts":1599907391592,"cs":"yyZsps7Cld3sEzvSzh7uIw==","size":{"width":258,"height":89}}

{"font":{"family":"Arial","size":12,"color":"#000000"},"type":"$","id":"5-0-1-0-1-0-1-0-1-0-1-1-1-1-0-1-1-0","code":"$16.\\,\\int_{}^{}\\left(2x-3\\cos x+e^{x}\\right)dx$","ts":1599907603021,"cs":"TvFaRYpqS73Z4GCby8CXwg==","size":{"width":232,"height":22}}

Solun:- Let f(x) = 2x - 3 cosx + ex

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(2x-3\\cos x+e^{x}\\right)dx}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"6-1-1-1-1-1-1-1-1-1-1-0","ts":1599907915797,"cs":"goB7LdhNwdO6pMlFSw11tw==","size":{"width":248,"height":36}}

We know that 

{"code":"\\begin{align*}\n{\\int_{}^{}x^{n}dx\\,\\,}&={\\frac{x^{n+1}}{n+1}+c}\\\\\n{\\int_{}^{}\\cos xdx}&={\\sin x+c}\\\\\n{\\int_{}^{}e^{x}dx\\,\\,\\,\\,\\,\\,}&={e^{x}+c}\t\n\\end{align*}","id":"7-1-1-1-1-1-1-1-3-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1599908007406,"cs":"+8e6tKbrlqDl2ubGS6GRcQ==","size":{"width":160,"height":118}}

{"type":"align*","id":"8-1-1-1-1-1-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{2x^{2}}{2}-3\\sin x+e^{x}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={x^{2}-3\\sin x+e^{x}+C}\t\n\\end{align*}","ts":1599908103270,"cs":"tAk48JLRsf07m2HE2aN1DA==","size":{"width":244,"height":78}}

{"id":"5-0-1-0-1-0-1-0-1-0-1-1-1-1-0-1-1-1-0","font":{"color":"#000000","family":"Arial","size":12},"code":"$17.\\,\\int_{}^{}\\left(2x^{2}-3\\sin x+5{\\sqrt[]{x}}\\right)dx$","type":"$","ts":1599908277259,"cs":"L4sT74geexyZQBc1AJD81g==","size":{"width":260,"height":24}}

Solun:- Let f(x) = {"font":{"family":"Arial","color":"#000000","size":11},"code":"$2x^{2}-3\\sin x+5{\\sqrt[]{x}}$","id":"11-0","type":"$","ts":1599908347625,"cs":"4yQdQgZu4jc+MWS0IiR6Bw==","size":{"width":152,"height":18}}

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(2x^{2}-3\\sin x+5{\\sqrt[]{x}}\\right)dx}\t\n\\end{align*}","id":"6-1-1-1-1-1-1-1-1-1-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1599908423796,"cs":"DdMJEPeHO/8dSunL4X2Sgw==","size":{"width":270,"height":36}}

We know that 

{"id":"7-1-1-1-1-1-1-1-3-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}x^{n}dx\\,\\,}&={\\frac{x^{n+1}}{n+1}+c}\\\\\n{\\int_{}^{}\\sin xdx}&={-\\cos x+c}\t\n\\end{align*}","type":"align*","ts":1599908745056,"cs":"dmW3+p9b99SDq7vx+wA5xw==","size":{"width":162,"height":78}}

{"id":"8-1-1-1-1-1-1-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{2x^{3}}{3}+3\\cos x+5\\frac{x^{\\frac{3}{2}}}{\\frac{3}{2}}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{2x^{3}}{3}+3\\cos x+\\frac{10x^{\\frac{3}{2}}}{3}+C}\t\n\\end{align*}","ts":1599908695323,"cs":"6/BGDy5TCXPtwO41xe61yQ==","size":{"width":272,"height":89}}

{"id":"5-0-1-0-1-0-1-0-1-0-1-1-1-1-0-1-1-1-1-0-0","type":"$","font":{"family":"Arial","color":"#000000","size":12},"code":"$18.\\,\\int_{}^{}\\sec x.\\left(\\sec x+\\tan x\\right)dx$","ts":1599908825450,"cs":"KIi+AyB2lz9S7QTFpNzW2w==","size":{"width":249,"height":22}}

Solun:- Let f(x) = {"code":"$\\sec x.\\left(\\sec x+\\tan x\\right)$","id":"5-0-1-0-1-0-1-0-1-0-1-1-1-1-0-1-1-1-1-1-0","font":{"size":11,"family":"Arial","color":"#000000"},"type":"$","ts":1599908879634,"cs":"pczlLbqzP4L3HK+PbB47xg==","size":{"width":156,"height":17}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sec x.\\left(\\sec x+\\tan x\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\sec ^{2}x+\\sec x.\\tan x\\right)dx}\t\n\\end{align*}","id":"6-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1599908972173,"cs":"2s+pqUkp933zmqN/RTnCyg==","size":{"width":268,"height":76}}

We know that 

{"font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\sec ^{2}x.dx}&={\\tan x+c}\t\n\\end{align*}","type":"align*","id":"7-1-1-1-1-1-1-1-3-1-1-0","ts":1599909033800,"cs":"qfbIXchTJvD/0utSj0PWeA==","size":{"width":164,"height":36}}

{"id":"7-1-1-1-1-1-1-1-3-1-1-1","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\sec x.\\tan x.dx}&={\\sec x+c}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1599909378194,"cs":"kdl/nH+9XzI76C1RTK1gqg==","size":{"width":196,"height":36}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\tan x+\\sec x+C}\t\n\\end{align*}","id":"8-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1599909471295,"cs":"fFAe41SydwdYac3ZyAwtaQ==","size":{"width":204,"height":36}}

{"code":"$19.\\,\\int_{}^{}\\frac{\\sec ^{2}x}{\\cos ec^{2}x}dx$","font":{"size":12,"color":"#000000","family":"Arial"},"type":"$","id":"5-0-1-0-1-0-1-0-1-0-1-1-1-1-0-1-1-1-1-0-1-0","ts":1599909624887,"cs":"Q7h5QOhNtQWV9epLbmTtdA==","size":{"width":130,"height":28}}

Solun:- Let f(x) = {"type":"$","code":"$\\frac{\\sec ^{2}x}{\\cos ec^{2}x}$","font":{"family":"Arial","size":11,"color":"#000000"},"id":"5-0-1-0-1-0-1-0-1-0-1-1-1-1-0-1-1-1-1-1-1-0","ts":1599909669395,"cs":"DecGzhoYub1KqLbADjCPHw==","size":{"width":50,"height":24}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"6-1-1-1-1-1-1-1-1-1-1-1-1-1-0-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\sec ^{2}x}{\\cos ec^{2}x}dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{ \\sin^{2}x}{\\cos^{2}x}dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\tan^{2}xdx}\t\n\\end{align*}","ts":1599909835887,"cs":"MGM4frVfvHVdDNshu+3Skg==","size":{"width":184,"height":121}}

We know that 

sec2x - tan2x = 1

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\sec^{2}x-1\\right)dx}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"id":"6-1-1-1-1-1-1-1-1-1-1-1-1-1-1","ts":1599909949038,"cs":"HSMgvQZ/mY8WlWNb/No9pA==","size":{"width":201,"height":36}}

We know that 

{"font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\sec ^{2}x.dx}&={\\tan x+c}\t\n\\end{align*}","type":"align*","id":"7-1-1-1-1-1-1-1-3-1-1-2","ts":1599909033800,"cs":"1kjQ64ozKmMTpotm8P9v9g==","size":{"width":164,"height":36}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}1.dx}&={x+c}\t\n\\end{align*}","id":"7-1-1-1-1-1-1-1-3-1-1-3","ts":1599909988658,"cs":"ipOOkHGYBnW5N7PttMp10Q==","size":{"width":104,"height":36}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\tan x-x+C}\t\n\\end{align*}","id":"8-1-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1599910264761,"cs":"UQt70wS3sQeYpFdKYZ6VLw==","size":{"width":181,"height":36}}

{"code":"$20.\\,\\int_{}^{}\\frac{2-3\\sin x}{\\cos ^{2}x}dx$","id":"5-0-1-0-1-0-1-0-1-0-1-1-1-1-0-1-1-1-1-0-1-1","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1599910382378,"cs":"xi/u7WSJ6sIIMWIihD2D2A==","size":{"width":137,"height":25}}

Solun:- Let f(x) = {"code":"$\\frac{2-3\\sin x}{\\cos ^{2}x}$","id":"5-0-1-0-1-0-1-0-1-0-1-1-1-1-0-1-1-1-1-1-1-1","type":"$","font":{"size":11,"family":"Arial","color":"#000000"},"ts":1599910417590,"cs":"jVFCXUBOoASIiZpkeUSN1w==","size":{"width":57,"height":22}}

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{2-3\\sin x}{\\cos ^{2}x}dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{ 2}{\\cos^{2}x}-\\frac{ 3\\sin x}{\\cos^{2}x}\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(2\\sec ^{2}x-3\\sec x.\\tan x\\right)dx}\t\n\\end{align*}","id":"6-1-1-1-1-1-1-1-1-1-1-1-1-1-0-1","ts":1599910773004,"cs":"1giXpkd+WsYntSG3SuEn2A==","size":{"width":288,"height":118}}

We know that 

{"font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}\\sec ^{2}x.dx}&={\\tan x+c}\t\n\\end{align*}","type":"align*","id":"7-1-1-1-1-1-1-1-3-1-1-0","ts":1599909033800,"cs":"qfbIXchTJvD/0utSj0PWeA==","size":{"width":164,"height":36}}

{"id":"7-1-1-1-1-1-1-1-3-1-1-1","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}\\sec x.\\tan x.dx}&={\\sec x+c}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1599909378194,"cs":"kdl/nH+9XzI76C1RTK1gqg==","size":{"width":196,"height":36}}

{"font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={2\\tan x-3\\sec x+C}\t\n\\end{align*}","type":"align*","id":"8-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1599910848391,"cs":"TwAXBjjW6984an4x1gjlyw==","size":{"width":225,"height":36}}

Choose the correct answer in Exercises 21 and 22.

21. The anti-derivative of {"code":"$\\left({\\sqrt[]{x}}+\\frac{1}{{\\sqrt[]{x}}}\\right)$","id":"12-0-0","type":"$","font":{"size":11,"family":"Arial","color":"#000000"},"ts":1599911228522,"cs":"adS/Ixiz6lQqpBPwRQd0Aw==","size":{"width":96,"height":32}} equals

{"id":"13","type":"gather*","font":{"family":"Arial","size":11,"color":"#000000"},"code":"\\begin{gather*}\n{\\left(A\\right)\\,\\frac{1}{3}x^{\\frac{1}{3}}+2x^{\\frac{1}{2}}+C\\,\\,}\\\\\n{\\left(B\\right)\\,\\frac{2}{3}x^{\\frac{2}{3}}+\\frac{1}{2}x^{2}+C\\,}\\\\\n{\\left(C\\right)\\,\\frac{2}{3}x^{\\frac{2}{3}}+2x^{\\frac{1}{2}}+C\\,\\,}\\\\\n{\\left(D\\right)\\,\\frac{3}{2}x^{\\frac{3}{2}}+\\frac{1}{2}x^{\\frac{1}{2}}+C}\t\n\\end{gather*}","ts":1599911526390,"cs":"8Ri+Ogz8r3U1fAvWSe1uCg==","size":{"width":166,"height":164}}

Solun:- Let f(x) = {"code":"$\\left({\\sqrt[]{x}}+\\frac{1}{{\\sqrt[]{x}}}\\right)$","font":{"color":"#000000","size":10,"family":"Arial"},"id":"12-1","type":"$","ts":1599912101902,"cs":"msJQWPpjG7gKV0JNr8Wg4A==","size":{"width":84,"height":28}}

{"font":{"size":10,"color":"#000000","family":"Arial"},"id":"14-0","code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}\\left({\\sqrt[]{x}}+\\frac{1}{{\\sqrt[]{x}}}\\right)dx$","type":"$","ts":1599912226500,"cs":"Hckowf0M0zRGJftVF+YnHQ==","size":{"width":194,"height":28}}

{"id":"14-1","type":"$","code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}\\left(x^{\\frac{1}{2}}+x^{\\frac{-1}{2}}\\right)dx$","font":{"family":"Arial","color":"#000000","size":10},"ts":1599912382409,"cs":"vx1i/xQaJbrjfydr6b9VzQ==","size":{"width":192,"height":28}}

We know that 

{"id":"7-1-1-1-1-1-1-1-3-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}x^{n}.dx}&={\\frac{x^{n+1}}{n+1}+c}\t\n\\end{align*}","type":"align*","ts":1599914173878,"cs":"owjUhsfyUErmcboh3JerHA==","size":{"width":150,"height":37}}

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"8-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x^{\\frac{3}{2}}}{\\frac{3}{2}}+\\frac{x^{\\frac{1}{2}}}{\\frac{1}{2}}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{2x^{\\frac{3}{2}}}{3}+2x^{\\frac{1}{2}}+C}\t\n\\end{align*}","ts":1599914385748,"cs":"Mg85HkLFH3Il9dsp9rTeWg==","size":{"width":198,"height":89}}

The correct answer is C.

22. If {"type":"$","font":{"size":11,"color":"#000000","family":"Arial"},"id":"12-0-1-0","code":"$\\diff{\\left(f\\left(x\\right)\\right)}{x}=4x^{3}-\\frac{3}{x^{4}}$","ts":1599914589828,"cs":"+h2iCqOMVYZPyLuocmPYXQ==","size":{"width":142,"height":25}} such that f(2) = 0. Then f(x) is

{"id":"13","code":"\\begin{gather*}\n{\\left(A\\right)\\,x^{4}+\\frac{1}{x^{3}}-\\frac{129}{8}\\,\\,}\\\\\n{\\left(B\\right)\\,x^{3}+\\frac{1}{x^{4}}+\\frac{129}{8}\\,\\,}\\\\\n{\\left(C\\right)\\,x^{4}+\\frac{1}{x^{3}}+\\frac{129}{8}\\,}\\\\\n{\\left(D\\right)\\,x^{3}+\\frac{1}{x^{4}}-\\frac{129}{8}}\t\n\\end{gather*}","font":{"color":"#000000","size":11,"family":"Arial"},"type":"gather*","ts":1599914848663,"cs":"/pVNanjCvsHpujZlxPKq0A==","size":{"width":160,"height":166}}

Solun:- Given {"type":"$","font":{"size":11,"color":"#000000","family":"Arial"},"id":"12-0-1-1","code":"$\\diff{\\left(f\\left(x\\right)\\right)}{x}=4x^{3}-\\frac{3}{x^{4}}$","ts":1599914589828,"cs":"pmbOG0X37ckqKu0JR0UiAQ==","size":{"width":142,"height":25}}

Anti-derivative of {"font":{"color":"#000000","size":10,"family":"Arial"},"id":"15","code":"$4x^{3}-\\frac{3}{x^{4}}$","type":"$","ts":1599915344546,"cs":"AkEtb7wGTk33W7YKESFe1Q==","size":{"width":60,"height":20}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","code":"\\begin{align*}\n{f\\left(x\\right)}&={\\int_{}^{}\\left(4x^{3}-\\frac{3}{x^{4}}\\right)dx}\\\\\n{f\\left(x\\right)}&={\\int_{}^{}\\left(4x^{3}-3x^{-4}\\right)dx}\t\n\\end{align*}","id":"14-2","ts":1599916081342,"cs":"IRsO589VsZ0D5HbZbVdyUw==","size":{"width":174,"height":77}}

We know that 

{"id":"7-1-1-1-1-1-1-1-3-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}x^{n}.dx}&={\\frac{x^{n+1}}{n+1}+c}\t\n\\end{align*}","type":"align*","ts":1599914173878,"cs":"owjUhsfyUErmcboh3JerHA==","size":{"width":150,"height":37}}

{"type":"align*","code":"\\begin{align*}\n{f\\left(x\\right)}&={\\frac{4x^{4}}{4}-\\frac{3x^{-3}}{-3}+C}\\\\\n{f\\left(x\\right)}&={x^{4}+\\frac{1}{x^{3}}+C}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"8-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1599916405654,"cs":"hB+n+v9M7xJEGTqEFy31qA==","size":{"width":170,"height":73}}...(1)

Calculate f(2):

{"id":"17","font":{"family":"Arial","size":9,"color":"#000000"},"code":"\\begin{align*}\n{f\\left(2\\right)}&={16+\\frac{1}{8}}\\\\\n{f\\left(2\\right)}&={\\frac{128+1}{8}+C}\\\\\n{f\\left(2\\right)}&={\\frac{129}{8}+C}\t\n\\end{align*}","type":"align*","ts":1599917139762,"cs":"mENvwefDTUC+U4l4gwsqOg==","size":{"width":138,"height":106}}

Given f(2) = 0

{"code":"\\begin{align*}\n{C}&={\\frac{-129}{8}}\t\n\\end{align*}","type":"align*","id":"18","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1599917301646,"cs":"J8oKO96FGwhgy9RQxJt0Qw==","size":{"width":76,"height":32}}

Put this value in Eq. 1:-

{"id":"8-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1","type":"align*","code":"\\begin{align*}\n{f\\left(x\\right)}&={x^{4}+\\frac{1}{x^{3}}-\\frac{129}{8}}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1599917551536,"cs":"UeJMkg5c4NcLhkLDsUhsyw==","size":{"width":158,"height":32}}

The correct answer is A.



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