12th Standard Mathematics — Applications Of Integration: Book Back MCQs with Answers & Explanations

Factor the constant out of the square root so the integral matches the standard inverse-sine pattern:

Hence

So the correct option is (A).

The absolute value changes definition at \(x=0\), so split the interval there. On \([-1,0]\) we have \(|x|=-x\); on \([0,2]\) we have \(|x|=x\):

Geometrically these are two triangles of area \(\tfrac12\) and \(2\). The answer is (C).

Apply the reflection property \( \int_{0}^{\pi} f(x)\,dx = \int_{0}^{\pi} f(\pi-x)\,dx \). Under \(x\mapsto \pi-x\):

• \(\cos^{2}(\pi-x)=\cos^{2}x\), so the exponential factor is unchanged;
• since \((2n+1)\) is odd, \(\cos\big[(2n+1)(\pi-x)\big]=-\cos\big[(2n+1)x\big]\), and cubing keeps the minus sign.

Therefore \(f(\pi-x) = -f(x)\), which forces

The answer is (C), independent of \(n\).

Check the symmetry first: with \(f(x)=\sin^{2}x\cos x\),

so \(f\) is even (a common trap is to read it as odd and answer 0). Using the even-function rule and then \(u=\sin x\):

The answer is (D).

The two arguments are reciprocals of each other, and for any \(x\neq 0\) both \(\dfrac{x^{2}}{x^{4}+1}\) and \(\dfrac{x^{4}+1}{x^{2}}\) are positive. Using the identity for positive \(t\),

the entire bracket simplifies to the constant \(\dfrac{\pi}{2}\). The integrand is constant, so

The answer is (D).

Write the integrand as \(\big(2x^{7}-3x^{5}+7x^{3}-x\big)\sec^{2}x + \sec^{2}x\). The factor \(\sec^{2}x\) is even, and each odd power of \(x\) makes its term an odd function, so all of those integrate to \(0\) over the symmetric interval. Only the constant term survives:

The answer is (C).

By the First Fundamental Theorem of Integral Calculus, differentiating a definite integral with respect to its upper limit simply returns the integrand evaluated at that limit:

Here \(g(t)=t\cos t\), so \(\dfrac{df}{dx}=x\cos x\) — there is no need to evaluate the integral itself. The answer is (C).

Comparing \(y^{2}=4x\) with \(y^{2}=4ax\) gives \(a=1\), so the latus rectum is the vertical line \(x=1\). The region is symmetric about the x-axis, so take twice the area of the upper half where \(y=2\sqrt{x}\):

The answer is (C).

Use the standard beta-type result

with \(m=1,\ n=99\):

The answer is (B).

Let \(I=\displaystyle\int_{0}^{\pi}\frac{dx}{1+5^{\cos x}}\). Replace \(x\) by \(\pi-x\); since \(\cos(\pi-x)=-\cos x\) and \(5^{-\cos x}=\dfrac{1}{5^{\cos x}}\),

Adding the two expressions for \(I\), the integrands combine to \(1\):

The answer is (A).

Apply the gamma recurrence \(\Gamma(k+1)=k\,\Gamma(k)\) twice:

Hence \(\dfrac{\Gamma(n+2)}{\Gamma(n)} = n(n+1) = 90\). Solving

the positive root is \(n=9\). The answer is (D).

Substitute \(u=3x\), so \(du=3\,dx\) and the limits become \(0\) to \(\pi/2\):

By the Wallis reduction, \(\displaystyle\int_{0}^{\pi/2}\cos^{3}u\,du=\frac{2}{3}\). Therefore

The answer is (B).

Since \(\sin^{4}x\) is symmetric about \(x=\dfrac{\pi}{2}\) on \([0,\pi]\),

The Wallis value is \(\displaystyle\int_{0}^{\pi/2}\sin^{4}x\,dx = \frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}=\frac{3\pi}{16}\). Doubling gives

The answer is (B).

This is a gamma integral. Using

with \(a=3\) and \(n=2\):

The answer is (D).

Using \(\displaystyle\int\frac{dx}{a^{2}+x^{2}}=\frac{1}{a}\tan^{-1}\frac{x}{a}\) with denominator \(4+x^{2}\) (so the parameter is \(2\)):

Set this equal to \(\dfrac{\pi}{8}\):

The answer is (D).

The curve meets the x-axis where \(y=0\), i.e. \(x(a-x)=0\), giving \(x=0\) and \(x=a\). Revolving about the x-axis,

The answer is (D).

In the second integral substitute \(t=x^{2}\), so \(dt=2x\,dx\) and \(\dfrac{dx}{x}=\dfrac{dt}{2t}\). The limits change from \(x=1,3\) to \(t=1,9\):

Comparing with \(\dfrac{1}{2}\big[f(a)-f(1)\big]\) gives \(a=9\). The answer is (C).

Put \(x=\sin\theta\), \(dx=\cos\theta\,d\theta\); the limits become \(0\) to \(\dfrac{\pi}{2}\):

Integrate by parts twice (taking \(\theta^{2}\) as the part to differentiate):

Evaluating from \(0\) to \(\dfrac{\pi}{2}\): at \(\dfrac{\pi}{2}\) we get \(\dfrac{\pi^{2}}{4}+0-2\), and at \(0\) we get \(0\). Hence

The answer is (D).

Use the trigonometric substitution \(x=a\sin\theta\), so \(\sqrt{a^{2}-x^{2}}=a\cos\theta\) and \(dx=a\cos\theta\,d\theta\). The limits become \(0\) to \(\dfrac{\pi}{2}\):

By Wallis, \(\displaystyle\int_{0}^{\pi/2}\cos^{4}\theta\,d\theta = \frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}=\frac{3\pi}{16}\), so the value is \(\dfrac{3\pi a^{4}}{16}\). The answer is (B).

Differentiate both sides with respect to \(x\). On the left, the Fundamental Theorem gives \(f(x)\). On the right, the derivative of \(\displaystyle\int_{x}^{1} t\,f(t)\,dt\) carries a minus sign (the variable is the lower limit), yielding \(-x f(x)\):

Putting \(x=1\) gives \(f(1)=\dfrac{1}{2}\). The answer is (A).