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Exercise 8.1 

1. Find the area of the region bounded by the curve y2 = x and the line x = 1, x = 4, and the x-axis in the first quadrant.

Solun:- Given eq. is:- y2 = x

⇒ y = √x

Area of given curve from x = 1 to x = 4 is:-

{"type":"align*","id":"1-0-0","font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{A}&={\\int_{a}^{b}y.dx}\\\\\n{A}&={\\int_{1}^{4}{\\sqrt[]{x}}.dx}\t\n\\end{align*}","ts":1603279138001,"cs":"s8yUFhjvC5qZ96yMRftx1A==","size":{"width":104,"height":82}}

We know that:-

{"font":{"color":"#222222","size":10,"family":"Arial"},"id":"2-0","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","type":"$","ts":1603279205673,"cs":"GX9Vr4oGjnp0ZIBBAXIc7Q==","size":{"width":136,"height":21}}

{"font":{"color":"#222222","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{A}&={\\left[\\frac{x^{\\frac{3}{2}}}{\\frac{3}{2}}\\right]_{1}^{4}}\\\\\n{A}&={\\frac{2}{3}\\left[x^{\\frac{3}{2}}\\right]_{1}^{4}}\\\\\n{A}&={\\frac{2}{3}\\left[\\left(4^{\\frac{3}{2}}\\right)-\\left(1^{\\frac{3}{2}}\\right)\\right]}\\\\\n{A}&={\\frac{2}{3}\\left[\\left(4{\\sqrt[]{4}}\\right)-\\left(1\\right)\\right]}\\\\\n{A}&={\\frac{14}{3}}\t\n\\end{align*}","id":"1-1-0","ts":1603279356123,"cs":"kT7yU6LZGwztAuUlk0GGqQ==","size":{"width":154,"height":200}}

2. Find the area of the region bounded by y2 = 9x, x = 2, x = 4, and the x-axis in the first quadrant.

Solun:- Given eq. is:- y2 = 9x

⇒ y = √9x = 3√x

Area of given curve from x = 2 to x = 4 is:-

{"id":"1-0-1","type":"align*","code":"\\begin{align*}\n{A}&={\\int_{a}^{b}y.dx}\\\\\n{A}&={\\int_{2}^{4}3{\\sqrt[]{x}}.dx}\t\n\\end{align*}","font":{"family":"Arial","color":"#222222","size":10},"ts":1603280803666,"cs":"1jBHNvEkwt47z12MHhOi/g==","size":{"width":112,"height":82}}

We know that:-

{"font":{"color":"#222222","size":10,"family":"Arial"},"id":"2-1","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","type":"$","ts":1603279205673,"cs":"Su31bZoev7Tf8bRllllkGw==","size":{"width":136,"height":21}}

{"font":{"size":10,"family":"Arial","color":"#222222"},"id":"1-1-1-0","type":"align*","code":"\\begin{align*}\n{A}&={3\\left[\\frac{x^{\\frac{3}{2}}}{\\frac{3}{2}}\\right]_{2}^{4}}\\\\\n{A}&={\\frac{6}{3}\\left[x^{\\frac{3}{2}}\\right]_{2}^{4}}\t\n\\end{align*}","ts":1603279825204,"cs":"WJFOqfHhF9uKdkoLlzkkbQ==","size":{"width":90,"height":88}}

{"font":{"color":"#222222","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{A}&={2\\left[\\left(4^{\\frac{3}{2}}\\right)-\\left(2^{\\frac{3}{2}}\\right)\\right]}\\\\\n{A}&={2\\left[\\left(4{\\sqrt[]{4}}\\right)-\\left(2{\\sqrt[]{2}}\\right)\\right]}\\\\\n{A}&={2\\left[8-\\left(2{\\sqrt[]{2}}\\right)\\right]}\\\\\n{A}&={16-4{\\sqrt[]{2}}}\t\n\\end{align*}","id":"3-0","ts":1603279913004,"cs":"mYnSugXPNMWVuFG9CT0sog==","size":{"width":172,"height":120}}

3. Find the area of the region bounded by x2 = 4y, y = 2, y = 4, and the y-axis in the first quadrant.

Solun:- Given eq. is:- x2 = 4y

⇒ x = √4y = 2√y

Area of the given curve from y = 2 to y = 4 is:-

{"font":{"color":"#222222","family":"Arial","size":10},"type":"align*","id":"1-0-2","code":"\\begin{align*}\n{A}&={\\int_{c}^{d}x.dy}\\\\\n{A}&={\\int_{2}^{4}2{\\sqrt[]{y}}.dy}\t\n\\end{align*}","ts":1603280781059,"cs":"DGoZSsHck1yqxBChedbebg==","size":{"width":109,"height":82}}

We know that:-

{"font":{"color":"#222222","size":10,"family":"Arial"},"id":"2-2","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","type":"$","ts":1603279205673,"cs":"4zkuJ+PN/by2yGmyy92kGQ==","size":{"width":136,"height":21}}

{"font":{"size":10,"color":"#222222","family":"Arial"},"type":"align*","id":"1-1-1-1-0","code":"\\begin{align*}\n{A}&={2\\left[\\frac{y^{\\frac{3}{2}}}{\\frac{3}{2}}\\right]_{2}^{4}}\\\\\n{A}&={\\frac{4}{3}\\left[y^{\\frac{3}{2}}\\right]_{2}^{4}}\t\n\\end{align*}","ts":1603280909993,"cs":"Db0UKrtBNBFUOtTDbrVb7g==","size":{"width":89,"height":88}}

{"code":"\\begin{align*}\n{A}&={\\frac{4}{3}\\left[\\left(4^{\\frac{3}{2}}\\right)-\\left(2^{\\frac{3}{2}}\\right)\\right]}\\\\\n{A}&={\\frac{4}{3}\\left[\\left(4{\\sqrt[]{4}}\\right)-\\left(2{\\sqrt[]{2}}\\right)\\right]}\\\\\n{A}&={\\frac{4}{3}\\left[8-\\left(2{\\sqrt[]{2}}\\right)\\right]}\\\\\n{A}&={\\frac{32-8{\\sqrt[]{2}}}{3}}\t\n\\end{align*}","type":"align*","id":"3-1-0","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603280980106,"cs":"Nj59wM7B2Bcoc/2pFkn39A==","size":{"width":178,"height":148}}

4. Find the area of the region bounded by the ellipse {"type":"$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"4","code":"$\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$","ts":1603281086477,"cs":"/F616Zuzq5qGYpkM/D6Wkw==","size":{"width":84,"height":22}}.

Solun:- Given eq. is:-

{"id":"5-0","code":"\\begin{align*}\n{\\frac{x^{2}}{16}+\\frac{y^{2}}{9}\\,\\,\\,\\,\\,\\,\\,\\,}&={1}\\\\\n{\\frac{x^{2}}{\\left(4\\right)^{2}}+\\frac{y^{2}}{\\left(3\\right)^{2}}}&={1}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","ts":1603281254287,"cs":"LdlJxjeV3qa0c+RfHqmANw==","size":{"width":116,"height":81}}

Area of given curve from x = 0 to x = 4 is:-

Area of given ellipse = 4(Area of region AOBA)

Area of region AOBA:-

{"font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{A}&={\\int_{a}^{b}y.dx}\\\\\n{A}&={\\int_{0}^{4}{\\sqrt[]{9\\left[1-\\frac{x^{2}}{16}\\right]}}.dx}\\\\\n{A}&={\\int_{0}^{4}{\\sqrt[]{9\\left[\\frac{16-x^{2}}{16}\\right]}}.dx}\\\\\n{A}&={\\frac{3}{4}\\int_{0}^{4}{\\sqrt[]{\\left(4\\right)^{2}-x^{2}}}.dx}\t\n\\end{align*}","id":"1-0-3","type":"align*","ts":1603284301641,"cs":"yKThN/mxRvDydZfBwf202A==","size":{"width":180,"height":189}}

We know that:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","code":"$\\int_{}^{}{\\sqrt[]{a^{2}-x^{2}}}.dx=\\frac{x}{2}{\\sqrt[]{a^{2}-x^{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{x}{a}\\right)+C$","id":"7","ts":1603276347627,"cs":"pmxiD2QbZs1LljHWa/mMsQ==","size":{"width":332,"height":20}}

{"id":"6-0","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{A}&={\\frac{3}{4}\\left[\\frac{x}{2}{\\sqrt[]{16-x^{2}}}+\\frac{16}{2}\\sin^{-1}\\left(\\frac{x}{4}\\right)\\right]_{0}^{4}}\\\\\n{A}&={\\frac{3}{4}\\left[\\left(\\frac{4}{2}{\\sqrt[]{16-16}}+\\frac{16}{2}\\sin^{-1}\\left(\\frac{4}{4}\\right)\\right)-\\left(\\frac{0}{2}{\\sqrt[]{16-0}}+\\frac{16}{2}\\sin^{-1}\\left(\\frac{0}{4}\\right)\\right)\\right]}\\\\\n{A}&={\\frac{3}{4}\\left[\\left(0+8\\sin^{-1}\\left(1\\right)\\right)-0\\right]}\\\\\n{A}&={\\frac{3}{4}\\times\\frac{8\\Pi}{2}}\\\\\n{A}&={3\\Pi }\t\n\\end{align*}","ts":1603354964792,"cs":"qnKm+l0408pLOdWz+ALZUg==","size":{"width":516,"height":178}}

Then Area of given ellipse = 4(3Ï€) = 12Ï€.

5. Find the area of the region bounded by the ellipse {"font":{"family":"Arial","color":"#000000","size":10},"code":"$\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$","id":"4","type":"$","ts":1603354574022,"cs":"VefaeLh+vaG0w6jshFP5uQ==","size":{"width":82,"height":22}}.

Solun:- Given eq. is:-

{"font":{"size":10,"family":"Arial","color":"#000000"},"id":"5-1","type":"align*","code":"\\begin{align*}\n{\\frac{x^{2}}{4}+\\frac{y^{2}}{9}\\,\\,\\,\\,\\,\\,\\,\\,}&={1}\\\\\n{\\frac{x^{2}}{\\left(2\\right)^{2}}+\\frac{y^{2}}{\\left(3\\right)^{2}}}&={1}\t\n\\end{align*}","ts":1603354601983,"cs":"1Zy8Ujs5QthJCwTQQLLOcA==","size":{"width":116,"height":81}}

Area of given curve from x = 0 to x = 2 is:-

Area of given ellipse = 4(Area of region AOBA)

Area of region AOBA:-

{"code":"\\begin{align*}\n{A}&={\\int_{a}^{b}y.dx}\\\\\n{A}&={\\int_{0}^{2}{\\sqrt[]{9\\left[1-\\frac{x^{2}}{4}\\right]}}.dx}\\\\\n{A}&={\\int_{0}^{2}{\\sqrt[]{9\\left[\\frac{4-x^{2}}{4}\\right]}}.dx}\\\\\n{A}&={\\frac{3}{2}\\int_{0}^{2}{\\sqrt[]{\\left(2\\right)^{2}-x^{2}}}.dx}\t\n\\end{align*}","id":"1-0-4-0","font":{"color":"#222222","family":"Arial","size":10},"type":"align*","ts":1603354721590,"cs":"TRvtgIN44EvGrHtG3r5eLg==","size":{"width":177,"height":189}}

We know that:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","code":"$\\int_{}^{}{\\sqrt[]{a^{2}-x^{2}}}.dx=\\frac{x}{2}{\\sqrt[]{a^{2}-x^{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{x}{a}\\right)+C$","id":"8","ts":1603276347627,"cs":"JVM3KcnA4DgcrkrwuIWvww==","size":{"width":332,"height":20}}

{"font":{"color":"#000000","family":"Arial","size":10},"id":"6-1-0","code":"\\begin{align*}\n{A}&={\\frac{3}{2}\\left[\\frac{x}{2}{\\sqrt[]{4-x^{2}}}+\\frac{4}{2}\\sin^{-1}\\left(\\frac{x}{2}\\right)\\right]_{0}^{2}}\\\\\n{A}&={\\frac{3}{2}\\left[\\left(\\frac{2}{2}{\\sqrt[]{4-4}}+\\frac{4}{2}\\sin^{-1}\\left(\\frac{2}{2}\\right)\\right)-\\left(\\frac{0}{2}{\\sqrt[]{4-0}}+\\frac{4}{2}\\sin^{-1}\\left(\\frac{0}{2}\\right)\\right)\\right]}\\\\\n{A}&={\\frac{3}{2}\\left[\\left(0+2\\sin^{-1}\\left(1\\right)\\right)-0\\right]}\\\\\n{A}&={\\frac{3}{2}\\times\\frac{2\\Pi}{2}}\\\\\n{A}&={\\frac{3\\Pi }{2}}\t\n\\end{align*}","type":"align*","ts":1603540396290,"cs":"yHKxt2kTqotmEwJyOpO16w==","size":{"width":476,"height":196}}

Then Area of given ellipse = 4(3Ï€/2) = 6Ï€.

6. Find the area of the region in the first quadrant enclosed by x-axis, line x = (√3)y, and the circle x2 + y2 = 4.

Solun:- Given eq. is:-

x2 + y2 = 4…....(1)

x = (√3)y…....(2)

Solving eq. 1 and 2:-

⇒ 3y2 + y2 = 4

⇒ 4y2 = 4

⇒ y2 = 1

⇒ y = 1,-1  and x = √3,-√3 

Area of given curve from x = 0 to x = 2 is:-

A = (Area of region from x = 0 to x = √3) +

       (Area of region from x = √3 to x = 2)

{"font":{"color":"#222222","family":"Arial","size":10},"id":"1-0-4-1","type":"align*","code":"\\begin{align*}\n{A}&={\\int_{c}^{d}y_{line}.dx+\\int_{a}^{b}y_{circle}.dx}\\\\\n{A}&={\\int_{0}^{{\\sqrt[]{3}}}\\frac{x}{{\\sqrt[]{3}}}.dx+\\int_{{\\sqrt[]{3}}}^{2}{\\sqrt[]{4-x^{2}}}.dx}\\\\\n{A}&={\\frac{1}{{\\sqrt[]{3}}}\\int_{0}^{{\\sqrt[]{3}}}x.dx+\\int_{{\\sqrt[]{3}}}^{2}{\\sqrt[]{4-x^{2}}}.dx}\t\n\\end{align*}","ts":1603358554783,"cs":"9HbzRhuU4OTw/2K15QSoGA==","size":{"width":261,"height":136}}

We know that:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","code":"$\\int_{}^{}{\\sqrt[]{a^{2}-x^{2}}}.dx=\\frac{x}{2}{\\sqrt[]{a^{2}-x^{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{x}{a}\\right)+C$","id":"9-0","ts":1603276347627,"cs":"wZvbHueEIYed2eiscvTjNQ==","size":{"width":332,"height":20}}

{"code":"\\begin{align*}\n{A}&={\\frac{1}{{\\sqrt[]{3}}}\\left[\\frac{x^{2}}{2}\\right]_{0}^{{\\sqrt[]{3}}}+\\left[\\frac{x}{2}{\\sqrt[]{4-x^{2}}}+\\frac{4}{2}\\sin^{-1}\\left(\\frac{x}{2}\\right)\\right]_{{\\sqrt[]{3}}}^{2}}\\\\\n{A}&={\\frac{1}{{\\sqrt[]{3}}}\\left[\\frac{3}{2}-0\\right]+\\left[\\left(\\frac{2}{2}{\\sqrt[]{4-4}}+\\frac{4}{2}\\sin^{-1}\\left(\\frac{2}{2}\\right)\\right)-\\left(\\frac{{\\sqrt[]{3}}}{2}{\\sqrt[]{4-3}}+\\frac{4}{2}\\sin^{-1}\\left(\\frac{{\\sqrt[]{3}}}{2}\\right)\\right)\\right]}\\\\\n{A}&={\\frac{{\\sqrt[]{3}}}{2}+\\left[\\left(0+2\\sin^{-1}\\left(1\\right)\\right)-\\left(\\frac{{\\sqrt[]{3}}}{2}+2\\sin^{-1}\\left(\\frac{{\\sqrt[]{3}}}{2}\\right)\\right)\\right]}\\\\\n{A}&={\\frac{{\\sqrt[]{3}}}{2}+\\left[\\left(0+\\frac{2\\Pi}{2}\\right)-\\left(\\frac{{\\sqrt[]{3}}}{2}+\\frac{2\\Pi}{3}\\right)\\right]}\\\\\n{A}&={\\frac{{\\sqrt[]{3}}}{2}+\\left[\\Pi-\\frac{{\\sqrt[]{3}}}{2}-\\frac{2\\Pi}{3}\\right]}\\\\\n{A}&={\\frac{\\Pi}{3}}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","id":"6-1-1-0","ts":1603364515180,"cs":"Cm17pkXZ058ZQXGORVTF0A==","size":{"width":600,"height":292}}

7. Find the area of smaller part of circle x2 + y2 = a2 cut off by line x = a/√2.

Solun:- Given eq. is:-

x2 + y2 = a2…....(1)

x = a/√2

Area of given curve from x = a/√2 to x = a is:-

A = 2(Area of region ADCA)

Area of region ADCA:-

{"id":"1-0-4-2","type":"align*","font":{"color":"#222222","size":10,"family":"Arial"},"code":"\\begin{align*}\n{A}&={\\int_{a}^{b}y.dx}\\\\\n{A}&={\\int_{\\frac{a}{{\\sqrt[]{2}}}}^{a}{\\sqrt[]{a^{2}-x^{2}}}.dx}\t\n\\end{align*}","ts":1603366712334,"cs":"QwllehH+s9KEpEXSZam9Ag==","size":{"width":150,"height":86}}

We know that:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","code":"$\\int_{}^{}{\\sqrt[]{a^{2}-x^{2}}}.dx=\\frac{x}{2}{\\sqrt[]{a^{2}-x^{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{x}{a}\\right)+C$","id":"9-1-0-0","ts":1603276347627,"cs":"ovXI9NPRj8GrDeadPll/0A==","size":{"width":332,"height":20}}

{"id":"11-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{}&={\\left[\\frac{x}{2}{\\sqrt[]{a^{2}-x^{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{x}{a}\\right)\\right]_{\\frac{a}{{\\sqrt[]{2}}}}^{a}}\\\\\n{}&={\\left[\\left(\\frac{a}{2}{\\sqrt[]{a^{2}-a^{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{a}{a}\\right)\\right)-\\left(\\frac{\\frac{a}{{\\sqrt[]{2}}}}{2}{\\sqrt[]{a^{2}-\\frac{a^{2}}{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{\\frac{a}{{\\sqrt[]{2}}}}{a}\\right)\\right)\\right]}\\\\\n{}&={\\left[\\left(\\frac{a}{2}\\times0+\\frac{a^{2}}{2}\\times\\frac{\\Pi}{2}\\right)-\\left(\\frac{a}{2{\\sqrt[]{2}}}{\\sqrt[]{\\frac{a^{2}}{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{1}{{\\sqrt[]{2}}}\\right)\\right)\\right]}\t\n\\end{align*}","ts":1603367389120,"cs":"2vHT9uaC5V3Tb4wlQKrMGA==","size":{"width":533,"height":150}}

{"id":"6-1-1-1","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{}&={\\left[\\left(\\frac{\\Pi a^{2}}{4}\\right)-\\left(\\frac{a}{2{\\sqrt[]{2}}}\\times\\frac{a}{{\\sqrt[]{2}}}+\\frac{a^{2}}{2}\\times\\frac{\\Pi}{4}\\right)\\right]}\\\\\n{}&={\\left[\\frac{\\Pi a^{2}}{4}-\\left(\\frac{a^{2}}{4}+\\frac{\\Pi a^{2}}{8}\\right)\\right]}\\\\\n{}&={\\frac{\\Pi a^{2}}{4}-\\frac{a^{2}}{4}-\\frac{\\Pi a^{2}}{8}}\\\\\n{}&={\\frac{\\Pi a^{2}}{8}-\\frac{a^{2}}{4}}\t\n\\end{align*}","ts":1603367410683,"cs":"KiX0lXRGmCjPs077522Xeg==","size":{"width":300,"height":164}}

{"id":"10","type":"align*","font":{"family":"Arial","color":"#222222","size":10},"code":"\\begin{align*}\n{A}&={2\\left(\\frac{\\Pi a^{2}}{8}-\\frac{a^{2}}{4}\\right)}\\\\\n{A}&={\\frac{\\Pi a^{2}}{4}-\\frac{a^{2}}{2}}\\\\\n{A}&={\\frac{a^{2}}{2}\\left(\\frac{\\Pi }{2}-1\\right)}\t\n\\end{align*}","ts":1603367317389,"cs":"Ov56+mouzAhP4MBkMzuLNQ==","size":{"width":138,"height":121}}

8. The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value of a.

Solun:- Given eq. is:-

y = √x…....(1)

Given:-

Area of region AEOCA = Area of region ACDBFEA

2(Area of region ACOA) = 2(Area of region ABDCA)

Area of region ACOA = Area of region ABDCA…...(1)

Area of region ACOA:-

{"code":"\\begin{align*}\n{A}&={\\int_{a}^{b}y.dx}\t\n\\end{align*}","id":"1-0-4-3-0","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","ts":1603445342792,"cs":"S1JT/mUHs0RBzRwI37NP0Q==","size":{"width":88,"height":38}}

{"font":{"size":10,"color":"#222222","family":"Arial"},"id":"12-0-0","type":"align*","code":"\\begin{align*}\n{}&={\\int_{0}^{a}{\\sqrt[]{x}}.dx}\t\n\\end{align*}","ts":1603446316223,"cs":"Qu4eCPavxA96HtajctNcKw==","size":{"width":92,"height":36}}

We know that:-

{"id":"9-1-1-0","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1603445206323,"cs":"jUZhqeLszDp1t0JrJpVxiQ==","size":{"width":136,"height":21}}

{"type":"align*","code":"\\begin{align*}\n{}&={\\left[\\frac{x^{\\frac{3}{2}}}{\\frac{3}{2}}\\right]_{0}^{a}}\\\\\n{}&={\\frac{2}{3}\\left[\\left(a^{\\frac{3}{2}}\\right)-0\\right]}\\\\\n{}&={\\frac{2a^{\\frac{3}{2}}}{3}}\t\n\\end{align*}","id":"11-1-0-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1603446377174,"cs":"fLRNsBU1nqKyofB5vs5+cg==","size":{"width":114,"height":128}}

Area of region ABDCA:-

{"id":"12-1","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","code":"\\begin{align*}\n{}&={\\int_{a}^{4}{\\sqrt[]{x}}.dx}\t\n\\end{align*}","ts":1603446397170,"cs":"NpS60PNpqFnr7/gjzhxS5Q==","size":{"width":92,"height":38}}

We know that:-

{"id":"9-1-1-1","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1603445206323,"cs":"J7DhOqP2lpp+wzwdGRS96w==","size":{"width":136,"height":21}}

{"type":"align*","id":"11-1-1","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{}&={\\left[\\frac{x^{\\frac{3}{2}}}{\\frac{3}{2}}\\right]_{a}^{4}}\\\\\n{}&={\\frac{2\\times4^{\\frac{3}{2}}}{3}-\\frac{2}{3}a^{\\frac{3}{2}}}\t\n\\end{align*}","ts":1603446525494,"cs":"hFSabwwCMRtVGSZ1AqPc/Q==","size":{"width":124,"height":93}}

From eq. 1:-

{"type":"align*","id":"13","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\frac{2a^{\\frac{3}{2}}}{3}}&={\\frac{2\\times4^{\\frac{3}{2}}}{3}-\\frac{2a^{\\frac{3}{2}}}{3}}\\\\\n{a^{\\frac{3}{2}}\\,\\,\\,\\,\\,}&={4^{\\frac{3}{2}}-a^{\\frac{3}{2}}}\\\\\n{2a^{\\frac{3}{2}}\\,\\,}&={4^{\\frac{3}{2}}}\\\\\n{a^{\\frac{3}{2}}\\,\\,\\,\\,\\,}&={\\frac{4^{\\frac{3}{2}}}{2}}\t\n\\end{align*}","ts":1603446938952,"cs":"QVLjbNxl6hvEemUlU5ByEw==","size":{"width":157,"height":128}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"14","code":"\\begin{align*}\n{a}&={\\left(\\frac{4^{\\frac{3}{2}}}{2}\\right)^{\\frac{2}{3}}}\\\\\n{a}&={\\frac{4^{\\frac{3}{2}\\times\\frac{2}{3}}}{2^{\\frac{2}{3}}}}\\\\\n{a}&={\\frac{4}{4^{\\frac{1}{3}}}}\\\\\n{a}&={4^{\\frac{2}{3}}}\t\n\\end{align*}","type":"align*","ts":1603447051574,"cs":"qRBipJTXyp0zAK14jYTRyQ==","size":{"width":88,"height":164}}

9. Find the area of the region bounded by the parabola y = x2 and y = |x|.

Solun:- Given eq. is:-

y = x2…....(1)

y = |x|

If x > 0 then y = x…....(2)

If x < 0 then y = -x…....(3)

From eq. 1 and 2:-

⇒ x = x2

⇒ x2 - x = 0

⇒ x(x - 1) = 0

⇒ x = 0 and y = 0 and

⇒ x = 1 and y = 1

From eq. 1 and 3:-

⇒ x = 0 and y = 0 and

⇒ x = -1 and y = 1

Area of bounded region between (y = x2) and (y = |x|) = 2(Area of shaded region)....(1)

Area of region ADOA = (area of ΔACO) - area of region ADOCA…....(2)

Area of ΔACO = 1/2

Area of region ADOCA:-

{"code":"\\begin{align*}\n{A}&={\\int_{a}^{b}y.dx}\t\n\\end{align*}","id":"1-0-4-3-1","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","ts":1603445342792,"cs":"hMycQWvq+tH85Lj1Xr2ejg==","size":{"width":88,"height":38}}

{"id":"12-0-1","code":"\\begin{align*}\n{}&={\\int_{0}^{1}x^{2}.dx}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1603449996011,"cs":"HlzLUkq7LtS+JA3WboaKpw==","size":{"width":85,"height":38}}

We know that:-

{"id":"9-1-1-2","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1603445206323,"cs":"d3gFGj0khFpBHpr/7EovMA==","size":{"width":136,"height":21}}

{"font":{"family":"Arial","color":"#000000","size":10},"id":"11-1-0-1-0","type":"align*","code":"\\begin{align*}\n{}&={\\left[\\frac{x^{3}}{3}\\right]_{0}^{1}}\\\\\n{}&={\\left[\\frac{1}{3}-0\\right]}\\\\\n{}&={\\frac{1}{3}}\t\n\\end{align*}","ts":1603450030551,"cs":"pnxBQ1NFxXuYcg+sAdaCbg==","size":{"width":80,"height":122}}

From eq. 2:-

Area of region ADOA = 1/2 - 1/3 = 1/6

From eq. 1:-

Area of bounded region between (y = x2) and (y = |x|) = 2(1/6)

Area of bounded region between (y = x2) and (y = |x|) = 1/3

10. Find the area bounded by the curve x2 = 4y and the line x = 4y - 2.

Solun:- Given eq. is:-

x2 = 4y…....(1)

x = 4y - 2…....(2)

From eq. 1 and 2:-

⇒ x = -1 and y = 1/4 and

⇒ x = 2 and y = 1

Area of shaded region = area(AEOA)+area(EBOE)...(1)

Area of region AEOA = area(ACOEA)-area(ACOA)

Area of AEOA:-

{"code":"\\begin{align*}\n{A}&={\\int_{a}^{b}\\left[y_{line}-y_{parabola}\\right].dx}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","id":"1-0-4-3-2","ts":1603452697464,"cs":"2dnpofG/WQ+vaOYT3zEOkQ==","size":{"width":184,"height":38}}

{"font":{"color":"#222222","family":"Arial","size":10},"code":"\\begin{align*}\n{}&={\\int_{0}^{2}\\left[\\frac{x+2}{4}-\\frac{x^{2}}{4}\\right].dx}\\\\\n{}&={\\int_{0}^{2}\\frac{x+2}{4}.dx-\\int_{0}^{2}\\frac{x^{2}}{4}.dx}\\\\\n{}&={\\frac{1}{4}\\int_{0}^{2}\\left(x+2\\right).dx-\\frac{1}{4}\\int_{0}^{2}x^{2}.dx}\t\n\\end{align*}","type":"align*","id":"12-0-2-0-0","ts":1603452809386,"cs":"n96i1g9wQcU6LGlyq8FoWw==","size":{"width":236,"height":126}}

We know that:-

{"id":"9-1-1-3","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1603445206323,"cs":"u++wTFQjWNvoWPhsnXW0UA==","size":{"width":136,"height":21}}

{"id":"11-1-0-1-1-0-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{}&={\\frac{1}{4}\\left[\\frac{x^{2}}{2}+2x\\right]_{0}^{2}-\\frac{1}{4}\\left[\\frac{x^{3}}{3}\\right]_{0}^{2}}\\\\\n{}&={\\frac{1}{4}\\left[\\left(\\frac{4}{2}+4\\right)-0\\right]-\\frac{1}{4}\\left[\\left(\\frac{8}{3}\\right)-0\\right]}\\\\\n{}&={\\frac{6}{4}-\\frac{2}{3}}\\\\\n{}&={\\frac{3}{2}-\\frac{2}{3}=\\frac{5}{6}}\t\n\\end{align*}","ts":1603452994625,"cs":"trsVgei5v1evNGuKqgKyYg==","size":{"width":261,"height":160}}

Area of EBOE:-

{"code":"\\begin{align*}\n{A}&={\\int_{a}^{b}\\left[y_{line}-y_{parabola}\\right].dx}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","id":"1-0-4-3-3","ts":1603452697464,"cs":"RhsfYHrco+GvRGZZTx2Row==","size":{"width":184,"height":38}}

{"font":{"family":"Arial","size":10,"color":"#222222"},"id":"12-0-2-1","type":"align*","code":"\\begin{align*}\n{}&={\\left|\\int_{-1}^{0}\\left[\\frac{x+2}{4}-\\frac{x^{2}}{4}\\right].dx\\right|}\\\\\n{}&={\\left|\\int_{-1}^{0}\\frac{x+2}{4}.dx-\\int_{-1}^{0}\\frac{x^{2}}{4}.dx\\right|}\\\\\n{}&={\\left|\\frac{1}{4}\\int_{-1}^{0}\\left(x+2\\right).dx-\\frac{1}{4}\\int_{-1}^{0}x^{2}.dx\\right|}\t\n\\end{align*}","ts":1603453119417,"cs":"9csaQxd5DIf8EEIKGyrlRw==","size":{"width":248,"height":133}}

We know that:-

{"id":"9-1-1-4","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1603445206323,"cs":"FZoIisnoAFcT34xZjQ4dRQ==","size":{"width":136,"height":21}}

{"type":"align*","id":"11-1-0-1-1-1","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{}&={\\left|\\frac{1}{4}\\left[\\frac{x^{2}}{2}+2x\\right]_{-1}^{0}-\\frac{1}{4}\\left[\\frac{x^{3}}{3}\\right]_{-1}^{0}\\right|}\\\\\n{}&={\\left|\\frac{1}{4}\\left[0-\\left(\\frac{1}{2}-2\\right)\\right]-\\frac{1}{4}\\left[0-\\left(\\frac{-1}{3}\\right)\\right]\\right|}\\\\\n{}&={\\left|\\frac{3}{8}-\\frac{1}{12}\\right|}\\\\\n{}&={\\left|\\frac{18-4}{48}\\right|=\\left|\\frac{14}{48}\\right|}\\\\\n{}&={\\frac{7}{24}}\t\n\\end{align*}","ts":1603544037543,"cs":"iabj/ebV+Z8RQoPkye3lGg==","size":{"width":284,"height":204}}

From eq. 1:-

Area of shaded region = 5/6 + 7/24 = 9/8

11. Find the area of the region bounded by the curve y2 = 4x and the line x = 3.

Solun:- Given eq. is:-

y2 = 4x

Area of region ABOA = 2(Area of region ACOA)....(1)

Area of region ACOA:-

{"font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{A}&={\\int_{a}^{b}y.dx}\t\n\\end{align*}","type":"align*","id":"1-0-4-3-2","ts":1603454238220,"cs":"N9YD4/cqYvsODLguKNN4JA==","size":{"width":88,"height":38}}

{"font":{"color":"#222222","size":10,"family":"Arial"},"type":"align*","id":"12-0-2-0-1-0","code":"\\begin{align*}\n{}&={\\int_{0}^{3}{\\sqrt[]{4x}}.dx}\\\\\n{}&={2\\int_{0}^{3}{\\sqrt[]{x}}.dx}\t\n\\end{align*}","ts":1603454288746,"cs":"0L7UHGfkwWH2o3liPrS2rA==","size":{"width":104,"height":82}}

We know that:-

{"id":"9-1-1-5","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1603445206323,"cs":"YSsJQPtnSeO23VAWyX7oDA==","size":{"width":136,"height":21}}

{"code":"\\begin{align*}\n{}&={2\\left[\\frac{x^{\\frac{3}{2}}}{\\frac{3}{2}}\\right]_{0}^{3}}\\\\\n{}&={\\frac{4}{3}\\left[\\left(3^{\\frac{3}{2}}\\right)-0\\right]}\\\\\n{}&={4{\\sqrt[]{3}}}\t\n\\end{align*}","type":"align*","id":"11-1-0-1-1-0-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1603454470586,"cs":"+GBHo7khviBCsb92zH4ojQ==","size":{"width":114,"height":112}}

From eq. 1:-

Area of ABOA = 2(4√3) = 8√3.

Choose the correct answer in the following Exercises 12 and 13.

12. Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2 is

(A) π (B) π/2 (C) π/3 (D) π/4

Solun:- Given eq. is:- x2 + y2 = 4

Area of the shaded region:-

{"font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{A}&={\\int_{a}^{b}y.dx}\t\n\\end{align*}","type":"align*","id":"1-0-4-3-2","ts":1603454238220,"cs":"N9YD4/cqYvsODLguKNN4JA==","size":{"width":88,"height":38}}

{"font":{"size":10,"family":"Arial","color":"#222222"},"id":"12-0-2-0-1-1-0","type":"align*","code":"\\begin{align*}\n{A}&={\\int_{0}^{2}{\\sqrt[]{4-x^{2}}}.dx}\t\n\\end{align*}","ts":1603457577174,"cs":"rYIxOz7Gn5kNPjAlj6JF3w==","size":{"width":140,"height":38}}

We know that:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","code":"$\\int_{}^{}{\\sqrt[]{a^{2}-x^{2}}}.dx=\\frac{x}{2}{\\sqrt[]{a^{2}-x^{2}}}+\\frac{a^{2}}{2}\\sin^{-1}\\left(\\frac{x}{a}\\right)+C$","id":"9-1-0-1-0","ts":1603276347627,"cs":"hYJonvvdrWIqyZz166ZVKg==","size":{"width":332,"height":20}}

{"type":"align*","code":"\\begin{align*}\n{A}&={\\left[\\frac{x}{2}{\\sqrt[]{4-x^{2}}}+\\frac{4}{2}\\sin^{-1}\\left(\\frac{x}{2}\\right)\\right]_{0}^{2}}\\\\\n{A}&={\\left[\\left(\\frac{2}{2}{\\sqrt[]{4-4}}+\\frac{4}{2}\\sin^{-1}\\left(\\frac{2}{2}\\right)\\right)-\\left(\\frac{0}{2}{\\sqrt[]{4-0}}+\\frac{4}{2}\\sin^{-1}\\left(\\frac{0}{2}\\right)\\right)\\right]}\\\\\n{A}&={\\left(0+\\frac{4}{2}\\sin^{-1}\\left(\\frac{2}{2}\\right)\\right)-0}\\\\\n{A}&={\\frac{2\\Pi}{2}=\\Pi}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"11-1-0-1-1-0-1-1","ts":1603457480663,"cs":"Zf8vK0640ttax8nu3N/big==","size":{"width":461,"height":164}}

The correct answer is A.

13. Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is

(A) 2 (B) 9/4 (C) 9/3 (D) 9/2

Solun:- Given eq. is:- y2 = 4x

Area of the shaded region:-


{"code":"\\begin{align*}\n{A}&={\\int_{a}^{b}x.dy}\t\n\\end{align*}","font":{"family":"Arial","color":"#222222","size":10},"id":"1-0-4-3-2","type":"align*","ts":1603518396296,"cs":"QNVNb/Ply6Bh/kc1Tvgacw==","size":{"width":88,"height":38}}

{"code":"\\begin{align*}\n{A}&={\\int_{0}^{3}\\frac{y^{2}}{4}.dy}\t\n\\end{align*}","type":"align*","font":{"color":"#222222","family":"Arial","size":10},"id":"12-0-2-0-1-1-1","ts":1603518447430,"cs":"pFyBwEJxelrHxX2b2ewyYw==","size":{"width":101,"height":38}}

We know that:-

{"id":"9-1-0-1-1","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","ts":1603518500814,"cs":"S1xtwoxufYC6x4yS7+R6xQ==","size":{"width":136,"height":21}}

{"code":"\\begin{align*}\n{A}&={\\frac{1}{4}\\left[\\frac{y^{3}}{3}\\right]_{0}^{3}}\\\\\n{A}&={\\frac{1}{4}\\left[\\left(\\frac{27}{3}\\right)-0\\right]}\\\\\n{A}&={\\frac{1}{4}\\times9}\\\\\n{A}&={\\frac{9}{4}}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"11-1-0-1-1-0-1-2","type":"align*","ts":1603518598097,"cs":"rGnLgeREDj7Mv1zRQZbD6Q==","size":{"width":137,"height":158}}

The correct answer is B.


Download PDF of Exercise 8.1


See Also:-

Notes of Application of Integrals


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