12th Standard Mathematics — Applications Of Integration: Online Practice Test

Explanation.

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

\[ \sqrt{4-9x^{2}} = 3\sqrt{\tfrac{4}{9}-x^{2}}. \]

Hence

\[ \int_{0}^{2/3}\frac{dx}{\sqrt{4-9x^{2}}} = \frac{1}{3}\int_{0}^{2/3}\frac{dx}{\sqrt{(2/3)^{2}-x^{2}}} = \frac{1}{3}\Big[\sin^{-1}\tfrac{x}{2/3}\Big]_{0}^{2/3} = \frac{1}{3}\sin^{-1}(1) = \frac{1}{3}\cdot\frac{\pi}{2} = \frac{\pi}{6}. \]

So the correct option is (A).

Explanation.

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 \(|x|=x\):

\[ \int_{-1}^{0}(-x)\,dx + \int_{0}^{2} x\,dx = \Big[-\tfrac{x^{2}}{2}\Big]_{-1}^{0} + \Big[\tfrac{x^{2}}{2}\Big]_{0}^{2} = \frac{1}{2} + 2 = \frac{5}{2}. \]

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

Explanation.

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

\[ I = \int_{0}^{\pi} f(x)\,dx = -\int_{0}^{\pi} f(x)\,dx \;\Rightarrow\; 2I = 0 \;\Rightarrow\; I = 0. \]

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

Explanation.

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

\[ f(-x)=\sin^{2}(-x)\cos(-x)=\sin^{2}x\cos x=f(x), \]

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

\[ I = 2\int_{0}^{\pi/2}\sin^{2}x\cos x\,dx = 2\int_{0}^{1} u^{2}\,du = 2\cdot\frac{1}{3} = \frac{2}{3}. \]

The answer is (D).

Explanation.

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 \(t\),

\[ \tan^{-1}t+\tan^{-1}\frac{1}{t}=\frac{\pi}{2}, \]

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

\[ I = \frac{\pi}{2}\times\big(4-(-4)\big) = \frac{\pi}{2}\times 8 = 4\pi. \]

The answer is (D).

Explanation.

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:

\[ \int_{-\pi/4}^{\pi/4}\sec^{2}x\,dx = 2\int_{0}^{\pi/4}\sec^{2}x\,dx = 2\big[\tan x\big]_{0}^{\pi/4} = 2(1-0) = 2. \]

The answer is (C).

Explanation.

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:

\[ \frac{d}{dx}\int_{0}^{x} g(t)\,dt = g(x). \]

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).

Explanation.

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 \(y=2\sqrt{x}\):

\[ A = 2\int_{0}^{1} 2\sqrt{x}\,dx = 4\Big[\tfrac{2}{3}x^{3/2}\Big]_{0}^{1} = \frac{8}{3}. \]

The answer is (C).

Explanation.

Use the standard beta-type result

\[ \int_{0}^{1} x^{m}(1-x)^{n}\,dx = \frac{m!\,n!}{(m+n+1)!}, \]

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

\[ \int_{0}^{1} x(1-x)^{99}\,dx = \frac{1!\,99!}{101!} = \frac{1}{101\times 100} = \frac{1}{10100}. \]

The answer is (B).

Explanation.

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

\[ I = \int_{0}^{\pi}\frac{dx}{1+5^{-\cos x}} = \int_{0}^{\pi}\frac{5^{\cos x}}{1+5^{\cos x}}\,dx. \]

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

\[ 2I = \int_{0}^{\pi}\frac{1+5^{\cos x}}{1+5^{\cos x}}\,dx = \int_{0}^{\pi} dx = \pi \;\Rightarrow\; I=\frac{\pi}{2}. \]

The answer is (A).

Explanation.

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

\[ \Gamma(n+2) = (n+1)\,\Gamma(n+1) = (n+1)\,n\,\Gamma(n). \]

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

\[ n^{2}+n-90 = 0 \;\Rightarrow\; (n-9)(n+10)=0, \]

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

Explanation.

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

\[ \int_{0}^{\pi/6}\cos^{3}3x\,dx = \frac{1}{3}\int_{0}^{\pi/2}\cos^{3}u\,du. \]

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

\[ \frac{1}{3}\cdot\frac{2}{3} = \frac{2}{9}. \]

The answer is (B).

Explanation.

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

\[ \int_{0}^{\pi}\sin^{4}x\,dx = 2\int_{0}^{\pi/2}\sin^{4}x\,dx. \]

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

\[ 2\cdot\frac{3\pi}{16} = \frac{3\pi}{8}. \]

The answer is (B).

Explanation.

This is a gamma integral. Using

\[ \int_{0}^{\infty} e^{-ax}\,x^{n}\,dx = \frac{n!}{a^{\,n+1}}, \]

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

\[ \int_{0}^{\infty} e^{-3x}x^{2}\,dx = \frac{2!}{3^{3}} = \frac{2}{27}. \]

The answer is (D).

Explanation.

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 \(2\)):

\[ \int_{0}^{a}\frac{dx}{4+x^{2}} = \frac{1}{2}\Big[\tan^{-1}\frac{x}{2}\Big]_{0}^{a} = \frac{1}{2}\tan^{-1}\frac{a}{2}. \]

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

\[ \frac{1}{2}\tan^{-1}\frac{a}{2} = \frac{\pi}{8} \;\Rightarrow\; \tan^{-1}\frac{a}{2} = \frac{\pi}{4} \;\Rightarrow\; \frac{a}{2}=1 \;\Rightarrow\; a=2. \]

The answer is (D).

Explanation.

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,

\[ V = \pi\int_{0}^{a} y^{2}\,dx = \pi\int_{0}^{a}\big(ax-x^{2}\big)\,dx = \pi\Big[\frac{ax^{2}}{2}-\frac{x^{3}}{3}\Big]_{0}^{a} = \pi a^{3}\Big(\frac{1}{2}-\frac{1}{3}\Big) = \frac{\pi a^{3}}{6}. \]

The answer is (D).

Explanation.

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 \(t=1,9\):

\[ \int_{1}^{3}\frac{e^{\sin x^{2}}}{x}\,dx = \frac{1}{2}\int_{1}^{9}\frac{e^{\sin t}}{t}\,dt = \frac{1}{2}\big[f(9)-f(1)\big]. \]

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

Explanation.

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

\[ I = \int_{0}^{\pi/2}\theta^{2}\cos\theta\,d\theta. \]

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

\[ \int \theta^{2}\cos\theta\,d\theta = \theta^{2}\sin\theta + 2\theta\cos\theta - 2\sin\theta. \]

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

\[ I = \frac{\pi^{2}}{4}-2. \]

The answer is (D).

Explanation.

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 \(\dfrac{\pi}{2}\):

\[ \int_{0}^{a}\big(a^{2}-x^{2}\big)^{3/2}\,dx = \int_{0}^{\pi/2} a^{3}\cos^{3}\theta\cdot a\cos\theta\,d\theta = a^{4}\int_{0}^{\pi/2}\cos^{4}\theta\,d\theta. \]

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).

Explanation.

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 \(-x f(x)\):

\[ f(x) = 1 - x\,f(x) \;\Rightarrow\; f(x)\,(1+x) = 1 \;\Rightarrow\; f(x) = \frac{1}{1+x}. \]

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