JEE Mathematics Definite Integration MCQs Set 38

Answer: (a) \( \dfrac32 \)

Explanation:
1. Using \( x\to9-x \) (since \( 3+6=9 \)), \( I=\int_3^6\dfrac{\sqrt{9-x}}{\sqrt x+\sqrt{9-x}}dx \).
2. Adding, \( 2I=\int_3^61\,dx=3 \), so \( I=\dfrac32 \). Answer: (a)

Teacher's Note:
1. Apply the reflection property with \( a+b=9 \).

Answer: (c) \( \pi\displaystyle\int_0^{\pi/2}f(\cos x)\,dx \)

Explanation:
1. Replacing \( x \) by \( \pi-x \) and adding gives \( 2I=\pi\int_0^\pi f(\sin x)dx \). Since \( \sin x \) is symmetric about \( \pi/2 \), \( I=\pi\int_0^{\pi/2}f(\sin x)\,dx \).
2. Replacing \( x \) by \( \dfrac\pi2-x \) turns this into \( \pi\int_0^{\pi/2}f(\cos x)\,dx \). Answer: (c)

Teacher's Note:
1. Use \( x\to\pi-x \) and then symmetry about \( \pi/2 \).

Answer: (a) \( [a]f(a)-\{f(1)+f(2)+\dots+f([a])\} \)

Explanation:
1. With \( n=[a] \), split the integral at \( 2,3,\dots,n \): \( \sum_{k=1}^{n-1}k\,[f(k+1)-f(k)]+n\,[f(a)-f(n)] \).
2. The sum telescopes to \( (n-1)f(n)-\sum_{j=1}^{n-1}f(j) \), so the total is \( nf(a)-\sum_{j=1}^nf(j)=[a]f(a)-\{f(1)+\dots+f([a])\} \). Answer: (a)

Teacher's Note:
1. On each interval \( [k,k+1) \) the greatest integer is the constant \( k \).

Answer: (a) \( 18 \)

Explanation:
1. The limit has the form \( \dfrac00 \), so apply L'Hopital's rule: the derivative of the numerator is \( 4[f(x)]^3f'(x) \).
2. The limit is \( 4\cdot6^3\cdot\dfrac1{48}=18 \). Answer: (a)

Teacher's Note:
1. Differentiate numerator and denominator.

Answer: (c) \( I_1\gt I_2 \)

Explanation:
1. For \( 0\lt x\lt1 \), \( x^3\lt x^2 \), so \( 2^{x^3}\lt2^{x^2} \) and \( I_2\lt I_1 \).
2. For \( x\gt1 \), \( x^3\gt x^2 \), so \( I_4\gt I_3 \). Hence only \( I_1\gt I_2 \) is correct. Answer: (c)

Teacher's Note:
1. Compare \( x^2 \) and \( x^3 \) on each interval.

Answer: (c) \( \dfrac\pi2 \)

Explanation:
1. Replace \( x \) by \( -x \): \( I=\int\dfrac{a^x\cos^2x}{1+a^x}dx \). Adding, \( 2I=\int_{-\pi}^\pi\cos^2x\,dx=\pi \).
2. So \( I=\dfrac\pi2 \). Answer: (c)

Teacher's Note:
1. Add the integral to its form with \( x\to-x \).

Answer: (c) \( 2 \)

Explanation:
1. Since \( 1+\sin2x=(\sin x+\cos x)^2 \) and \( \sin x+\cos x\gt0 \) on \( [0,\pi/2] \), the integrand is \( \sin x+\cos x \).
2. \( \int_0^{\pi/2}(\sin x+\cos x)dx=[-\cos x+\sin x]_0^{\pi/2}=1+1=2 \). Answer: (c)

Teacher's Note:
1. Simplify the radicand.

Answer: (a) \( 2 \)

Explanation:
1. \( f(a)+f(-a)=\dfrac{e^a}{1+e^a}+\dfrac1{1+e^a}=1 \), so the limits are \( t \) and \( 1-t \) with \( t=f(-a) \).
2. Using \( x\to1-x \) (the limits sum to 1): \( I_1=\int(1-x)g[x(1-x)]dx \). Adding, \( 2I_1=\int g[x(1-x)]dx=I_2 \), so \( \dfrac{I_2}{I_1}=2 \). Answer: (a)

Teacher's Note:
1. Show that the limits add up to 1 and reflect.

Answer: (b) \( e-1 \)

Explanation:
1. This is a Riemann sum for \( \int_0^1e^x\,dx \).
2. \( \int_0^1e^xdx=e-1 \). Answer: (b)

Teacher's Note:
1. Identify the Riemann sum.

Answer: (b) \( \dfrac{a+b}2\displaystyle\int_a^bf(x)\,dx \)

Explanation:
1. Replacing \( x \) by \( a+b-x \): \( I=\int_a^b(a+b-x)f(x)\,dx=(a+b)\int_a^bf-I \).
2. Hence \( I=\dfrac{a+b}2\int_a^bf(x)\,dx \). Answer: (b)

Teacher's Note:
1. Reflect the integral using the given symmetry.

Answer: (c) \( \dfrac1{n+1}-\dfrac1{n+2} \)

Explanation:
1. Replacing \( x \) by \( 1-x \): \( I=\int_0^1(1-x)x^n\,dx=\int_0^1(x^n-x^{n+1})\,dx \).
2. \( I=\dfrac1{n+1}-\dfrac1{n+2} \). Answer: (c)

Teacher's Note:
1. Use \( \int_0^1f(x)dx=\int_0^1f(1-x)dx \).

Answer: (d) \( \dfrac15 \)

Explanation:
1. The first limit is \( \int_0^1x^4dx=\dfrac15 \).
2. The second is \( \dfrac1n\cdot\dfrac1n\sum\left(\dfrac rn\right)^3\to0\cdot\dfrac14=0 \), so the difference is \( \dfrac15 \). Answer: (d)

Teacher's Note:
1. Write each sum as a Riemann sum.

Answer: (d) \( 64 \)

Explanation:
1. Put \( z=x^3 \), so \( 3x^2dx=dz \) and \( \dfrac3x\,dx=\dfrac{3x^2}{x^3}dx=\dfrac{dz}z \).
2. The integral becomes \( \int_1^{64}\dfrac{e^{\sin z}}z\,dz=F(64)-F(1) \), so \( k=64 \). Answer: (d)

Teacher's Note:
1. Substitute \( z=x^3 \).

Answer: (c) \( e-\dfrac{e^2}2-\dfrac32 \)

Explanation:
1. \( f(x)=e^x \) and \( g(x)=x^2-e^x \), so the integral is \( \int_0^1x^2e^xdx-\int_0^1e^{2x}dx \).
2. \( \int_0^1x^2e^xdx=e-2 \) and \( \int_0^1e^{2x}dx=\dfrac{e^2-1}2 \), so the value is \( e-2-\dfrac{e^2-1}2=e-\dfrac{e^2}2-\dfrac32 \). Answer: (c)

Teacher's Note:
1. Solve for \( f \) and \( g \) first.

Answer: (a) \( 20 \)

Explanation:
1. \( |\sin x| \) has period \( \pi \), so the integral is \( 10\int_0^\pi\sin x\,dx \).
2. \( \int_0^\pi\sin x\,dx=2 \), so the value is \( 20 \). Answer: (a)

Teacher's Note:
1. Use the period of \( |\sin x| \).

Answer: (b) \( 1 \)

Explanation:
1. \( I_n+I_{n+2}=\int_0^{\pi/4}\tan^nx\sec^2x\,dx=\int_0^1t^ndt=\dfrac1{n+1} \).
2. So \( n[I_n+I_{n+2}]=\dfrac n{n+1}\to1 \). Answer: (b)

Teacher's Note:
1. Combine \( I_n \) and \( I_{n+2} \) using \( 1+\tan^2x=\sec^2x \).

Answer: (d) \( -\sqrt2-\sqrt3+5 \)

Explanation:
1. \( [x^2] \) is \( 0 \) on \( [0,1) \), \( 1 \) on \( [1,\sqrt2) \), \( 2 \) on \( [\sqrt2,\sqrt3) \) and \( 3 \) on \( [\sqrt3,2) \).
2. Integral \( =(\sqrt2-1)+2(\sqrt3-\sqrt2)+3(2-\sqrt3)=5-\sqrt3-\sqrt2 \). Answer: (d)

Teacher's Note:
1. Find where \( x^2 \) crosses each integer.

Answer: (b) \( \pi^2 \)

Explanation:
1. Split: \( \dfrac{2x}{1+\cos^2x} \) is odd and integrates to \( 0 \); \( \dfrac{2x\sin x}{1+\cos^2x} \) is even, so \( I=4\int_0^\pi\dfrac{x\sin x}{1+\cos^2x}dx \).
2. Using \( x\to\pi-x \): \( \int_0^\pi\dfrac{x\sin x}{1+\cos^2x}dx=\dfrac\pi2\int_0^\pi\dfrac{\sin x}{1+\cos^2x}dx=\dfrac\pi2\cdot\dfrac\pi2 \), so \( I=4\cdot\dfrac{\pi^2}4=\pi^2 \). Answer: (b)

Teacher's Note:
1. Separate odd and even parts.

Answer: (a) \( \dfrac\pi4 \)

Explanation:
1. Replace \( x \) by \( \dfrac\pi2-x \): \( I=\int_0^{\pi/2}\dfrac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx \).
2. Adding, \( 2I=\int_0^{\pi/2}dx=\dfrac\pi2 \), so \( I=\dfrac\pi4 \). Answer: (a)

Teacher's Note:
1. Add the integral to its reflected form.