JEE Mathematics Application of Integrals MCQs Set 10

Answer: (c) \(\dfrac43\)

Explanation:
1. \(x^2+3=2x+3\) gives \(x=0,2\).
2. Area \(=\displaystyle\int_0^2(2x-x^2)dx=4-\dfrac83=\dfrac43\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(\left(9\sec^{-1}3-\sqrt8\right)\) sq. unit

Explanation:
1. Area \(=2\displaystyle\int_1^3\sqrt{9-x^2}\,dx=2\left[\dfrac x2\sqrt{9-x^2}+\dfrac92\sin^{-1}\dfrac x3\right]_1^3\).
2. \(=9\cdot\dfrac\pi2-\sqrt8-9\sin^{-1}\dfrac13=9\cos^{-1}\dfrac13-\sqrt8=9\sec^{-1}3-\sqrt8\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(\dfrac43\) sq. units

Explanation:
1. The line meets the parabola at \(x=\pm1\).
2. Area \(=2\displaystyle\int_0^1(1-x^2)dx=2\left(1-\dfrac13\right)=\dfrac43\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (b) \(8\) sq. units

Explanation:
1. The lines \(y=\pm2x\) meet \(y=4\) at \(x=\pm2\).
2. Triangle with base \(4\) and height \(4\): area \(=\dfrac12\cdot4\cdot4=8\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(-3\)

Explanation:
1. \(f'(1)=2+b\); tangent: \(y-1=(2+b)(x-1)\) with intercepts \(1-\dfrac1{2+b}\) and \(-(1+b)\).
2. Area \(=\dfrac12\left(1-\dfrac1{2+b}\right)(-1-b)=2\) gives \(b=-3\) (the tangent \(y=2-x\), intercepts \(2\) and \(2\)).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(\dfrac{14}9\) sq. units

Explanation:
1. Integrate with respect to \(y\): \(x=\dfrac{\sqrt y}3\).
2. Area \(=\displaystyle\int_1^4\dfrac{\sqrt y}3dy=\dfrac29\left(8-1\right)=\dfrac{14}9\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (c) \(1\)

Explanation:
1. \(A(a)=\displaystyle\int_a^{2a}\left(\dfrac x6+\dfrac1{x^2}\right)dx=\dfrac{a^2}4+\dfrac1{2a}\).
2. \(A'(a)=\dfrac a2-\dfrac1{2a^2}=0\) gives \(a^3=1\), \(a=1\) (a minimum, since \(A''>0\)).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(\dfrac13\{f(0)+4f(1)+f(2)\}\)

Explanation:
1. \(f>0\) everywhere, so area \(=\displaystyle\int_0^2(a+bx+cx^2)dx=2a+2b+\dfrac{8c}3\).
2. \(f(0)+4f(1)+f(2)=6a+6b+8c\), and \(\dfrac13\) of it equals \(2a+2b+\dfrac{8c}3\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(1\) sq. unit

Explanation:
1. The curve cuts the \(x\)-axis at \(x=1-e\) and the \(y\)-axis at \(y=1\).
2. Area \(=\displaystyle\int_{1-e}^0\ln(x+e)dx=\displaystyle\int_1^e\ln u\,du=\left[u\ln u-u\right]_1^e=1\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (c) \(x+1+\dfrac x{\sqrt{1+x^2}}\)

Explanation:
1. \(\displaystyle\int_1^t(f(x)-x)dx=t+\sqrt{1+t^2}-\sqrt2-1\).
2. Differentiating in \(t\): \(f(t)-t=1+\dfrac t{\sqrt{1+t^2}}\), so \(f(x)=x+1+\dfrac x{\sqrt{1+x^2}}\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(1\)

Explanation:
1. For \(c<1\): \(\displaystyle\int_c^1(8x^2-x^5)dx=\dfrac{8}3-\dfrac16-\dfrac{8c^3}3+\dfrac{c^6}6=\dfrac{16}3\).
2. This is satisfied by \(c=-1\) (and no other \(c<1\)), so \(c+2=1\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (b) \(\dfrac12\ln2\) sq. units

Explanation:
1. Area \(=\displaystyle\int_0^{\pi/4}\tan x\,dx=\left[-\ln\cos x\right]_0^{\pi/4}\).
2. \(=-\ln\dfrac1{\sqrt2}=\dfrac12\ln2\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (c) \(6\) sq. units

Explanation:
1. Integrate with respect to \(y\): \(x=\dfrac{y^2}{12}\).
2. Area \(=\displaystyle\int_0^6\dfrac{y^2}{12}dy=\dfrac{216}{36}=6\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(\left(e+\dfrac1e-2\right)\) sq. units

Explanation:
1. The curves meet at \(x=0\) and \(e^x\ge e^{-x}\) for \(x\ge0\).
2. Area \(=\displaystyle\int_0^1(e^x-e^{-x})dx=\left[e^x+e^{-x}\right]_0^1=e+\dfrac1e-2\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(4\) sq. units

Explanation:
1. \(y'=(3-x)(3-3x)=0\) gives the maximum at \(x=1\) and the minimum at \(x=3\).
2. Area \(=\displaystyle\int_1^3x(3-x)^2dx=\displaystyle\int_0^2(3-u)u^2du=8-4=4\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (b) \(\dfrac{3\pi ab}{32}\)

Explanation:
1. Area \(=\displaystyle\int y\,dx=\int_{\pi/2}^0b\sin^3t\cdot(-3a\cos^2t\sin t)dt=3ab\int_0^{\pi/2}\sin^4t\cos^2t\,dt\).
2. \(\displaystyle\int_0^{\pi/2}\sin^4t\cos^2t\,dt=\dfrac\pi{32}\), so the area is \(\dfrac{3\pi ab}{32}\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(\dfrac7{12}\) sq. units

Explanation:
1. The curves meet at \(x=0\) and \(x=1\) (since \(\ln x+2x-2=0\) has the single root \(x=1\)).
2. Area \(=\displaystyle\int_0^1(2x-2x^2-x\ln x)dx=1-\dfrac23+\dfrac14=\dfrac7{12}\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(\sin(3x+4)+3(x-1)\cos(3x+4)\)

Explanation:
1. \(\displaystyle\int_1^bf(x)dx=(b-1)\sin(3b+4)\).
2. Differentiating in \(b\) (product rule): \(f(b)=\sin(3b+4)+3(b-1)\cos(3b+4)\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (c) \(2\) sq. units

Explanation:
1. Area \(=\displaystyle\int_6^9\dfrac{36}{x^2}dx=36\left(\dfrac16-\dfrac19\right)\).
2. \(=36\cdot\dfrac1{18}=2\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(9\) sq. units

Explanation:
1. Slope at \((2,3)\): \(2(y-2)y'=1\) gives \(y'=\dfrac12\); tangent \(y=\dfrac x2+2\), i.e. \(x=2y-4\).
2. Area \(=\displaystyle\int_0^3\left[(y-2)^2+1-(2y-4)\right]dy=\int_0^3(y-3)^2dy=9\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(\pi a^2\) sq. units

Explanation:
1. \(x=\dfrac{a^3}{y^2+a^2}\).
2. Area \(=\displaystyle\int_{-\infty}^{\infty}\dfrac{a^3}{y^2+a^2}dy=a^3\cdot\dfrac\pi a=\pi a^2\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (c) \(4\) sq. units

Explanation:
1. For \(x>0\) the curves \(\ln x\) and \(|\ln x|\) coincide for \(x\ge1\) and enclose a region on \(0<x<1\): \(\displaystyle\int_0^1(-2\ln x)dx=2\).
2. By symmetry in the \(y\)-axis the left half contributes another \(2\), total \(4\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(2\sqrt2\)

Explanation:
1. Total area \(=\displaystyle\int_2^4\left(1+\dfrac8{x^2}\right)dx=2+8\left(\dfrac12-\dfrac14\right)=4\), so each half is \(2\).
2. \(\displaystyle\int_2^a\left(1+\dfrac8{x^2}\right)dx=a+2-\dfrac8a=2\) gives \(a^2=8\), \(a=2\sqrt2\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (b) \(2\)

Explanation:
1. \(x=\sin y\), so the area is \(\displaystyle\int_{-\pi/2}^{\pi/2}|\sin y|\,dy\).
2. \(=2\displaystyle\int_0^{\pi/2}\sin y\,dy=2\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (c) \(\dfrac56\) sq. units

Explanation:
1. Area \(=\displaystyle\int_0^1x^2dx+\int_1^2(2-x)dx\).
2. \(=\dfrac13+\dfrac12=\dfrac56\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(\dfrac43\) sq. units

Explanation:
1. \(-(x-1)(x-3)=0\) gives \(x=1,3\).
2. Area \(=\displaystyle\int_1^3(4x-x^2-3)dx=\dfrac{(3-1)^3}6=\dfrac43\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(2\pi\) sq. units

Explanation:
1. The radius is \(\sqrt2\).
2. Area \(=\pi r^2=2\pi\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (b) \(\dfrac1{\sqrt3}\)

Explanation:
1. The curves meet at \((0,0)\) and \(\left(\dfrac1a,\dfrac1a\right)\).
2. Area \(=\displaystyle\int_0^{1/a}\left(\sqrt{\dfrac xa}-ax^2\right)dx=\dfrac1{3a^2}=1\), so \(a=\dfrac1{\sqrt3}\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (d) \(-\dfrac34\)

Explanation:
1. Integrate by parts: \(\displaystyle\int_0^2xf'(x)dx=\left[xf(x)\right]_0^2-\int_0^2f(x)dx\).
2. \(f(2)=0\) so the bracket is \(0\), and \(\displaystyle\int_0^2f\,dx=\dfrac34\); the result is \(-\dfrac34\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.

Answer: (a) \(\dfrac56\) sq. unit

Explanation:
1. \(f(x)=x^2+x+c\) and \(f(1)=2\) gives \(c=0\), so \(f(x)=x^2+x\), which cuts the \(x\)-axis at \(x=0\).
2. Area \(=\displaystyle\int_0^1(x^2+x)dx=\dfrac13+\dfrac12=\dfrac56\).

Teacher's Note:
1. Sketch the region, find the points of intersection and integrate the upper curve minus the lower curve (or use symmetry).
2. Use area = integral of |f - g| and split the interval wherever the curves cross.