Q.Evaluate $\int_{-1}^1 5x^4 \sqrt{x^5 + 1}\, dx$

Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.$\displaystyle\int_{-1}^{4} \sqrt{\dfrac{4-x}{x+1}}\,dx =$ (A) 0 (B) $\dfrac{\pi}{2}$ (C) $\dfrac{3\pi}{2}$ (D) $\dfrac{5\pi}{2}$

   ›Reveal solutionSolution

   This integral matches the standard result $\int_a^b\sqrt{\frac{b-x}{x-a}}\,dx=\frac{\pi(b-a)}2$, giving $\frac{5\pi}2$.

   Concept and Intuition

   Integrals of the form $\int_a^b\sqrt{\dfrac{b-x}{x-a}}\,dx$ arise often and have a clean closed form obtained via the substitution $x=a+(b-a)\sin^2\theta$, which converts the square root into $\cot\theta$ and the whole integral into $\int_0^{\pi/2}\cos^2\theta\,d\theta$.

   Step-by-Step Solution

   1. Here $a=-1,\ b=4$ (matching $\sqrt{\frac{4-x}{x+1}}=\sqrt{\frac{b-x}{x-a}}$).
   2. Substitute $x=a+(b-a)\sin^2\theta$: then $b-x=(b-a)\cos^2\theta$, $x-a=(b-a)\sin^2\theta$, so $\sqrt{\frac{b-x}{x-a}}=\cot\theta$, and $dx=2(b-a)\sin\theta\cos\theta\,d\theta$.
   3. The integral becomes $\displaystyle\int_0^{\pi/2}\cot\theta\cdot2(b-a)\sin\theta\cos\theta\,d\theta=2(b-a)\int_0^{\pi/2}\cos^2\theta\,d\theta=2(b-a)\cdot\frac\pi4=\frac{\pi(b-a)}2$.
   4. With $b-a=4-(-1)=5$: integral $=\dfrac{5\pi}2$.

   Common Mistakes

   • Mixing up which endpoint is $a$ and which is $b$ (the numerator under the root, $4-x$, must vanish at the upper limit $x=4$).
   • Forgetting the factor of $2$ from $dx$ when substituting.

   ✓Final answer

   The correct option is (D) — $\dfrac{5\pi}{2}$.

   ANSWER: D

2. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$\displaystyle\int_0^1 \dfrac{x}{(1-x)^{3/4}}\,dx =$ (A) $\dfrac{4}{5}$ (B) $\dfrac{8}{15}$ (C) $\dfrac{14}{5}$ (D) $\dfrac{16}{5}$

   ›Reveal solutionSolution

   With $u=1-x$, the integral becomes $\int_0^1(u^{-3/4}-u^{1/4})du = \dfrac{16}{5}$.

   Concept and Intuition

   Integrals of the form $\int x^m(1-x)^n dx$ over $[0,1]$ are cleanly handled by substituting $u=1-x$ so the fractional power becomes a simple power of $u$.

   Step-by-Step Solution

   1. Let $u=1-x \Rightarrow x=1-u,\ dx=-du$; limits $x:0\to1$ becomes $u:1\to0$.
   2. $\displaystyle\int_0^1\dfrac{x}{(1-x)^{3/4}}dx = \int_0^1\dfrac{1-u}{u^{3/4}}du = \int_0^1\left(u^{-3/4}-u^{1/4}\right)du$.
   3. $= \left[4u^{1/4} - \dfrac{4}{5}u^{5/4}\right]_0^1 = 4-\dfrac{4}{5} = \dfrac{16}{5}$.

   Common Mistakes

   • Forgetting to flip the limits when substituting (or equivalently forgetting the sign from $dx=-du$) — here the two sign flips cancel, but it's easy to lose track and get the wrong sign overall.
   • Arithmetic slip on $\int u^{-3/4}du = 4u^{1/4}$ (many mistakenly divide by $-3/4$ instead of $1/4$).

   ✓Final answer

   The correct option is (D) — $\dfrac{16}{5}$.

   ANSWER: D

3. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.$\int_{1/\sqrt[5]{31}}^{1/\sqrt[5]{242}}\dfrac{1}{\sqrt[5]{x^{30}+x^{25}}}\,dx=$ (A) $\dfrac{65}{4}$ (B) $\dfrac{-75}{4}$ (C) $\dfrac{75}{4}$ (D) $\dfrac{-65}{4}$

   ›Reveal solutionSolution

   Factor out the highest power of $x$ inside the fifth root to expose a clean substitution $w=1+x^{-5}$, then evaluate at the given nasty-looking but designed-to-simplify limits.

   Concept and Intuition

   The limits $31^{-1/5}$ and $242^{-1/5}$ look intimidating, but they're chosen precisely so that $1+x^{-5}$ becomes the perfect fifth powers $32=2^5$ and $243=3^5$ at the two ends — a strong hint to substitute $w=1+x^{-5}$.

   Step-by-Step Solution

   1. $x^{30}+x^{25}=x^{25}(x^5+1)$, so $(x^{30}+x^{25})^{1/5}=x^5(x^5+1)^{1/5}=x^5\cdot x(1+x^{-5})^{1/5}=x^6(1+x^{-5})^{1/5}$ (for $x>0$).
   2. So the integrand is $\dfrac{1}{x^6(1+x^{-5})^{1/5}}=x^{-6}(1+x^{-5})^{-1/5}$.
   3. Let $w=1+x^{-5}$, so $dw=-5x^{-6}dx\Rightarrow x^{-6}dx=-\dfrac{dw}{5}$.
   4. $\displaystyle\int x^{-6}(1+x^{-5})^{-1/5}dx=-\frac15\int w^{-1/5}dw=-\frac15\cdot\frac{w^{4/5}}{4/5}+c=-\frac14w^{4/5}+c$.
   5. Lower limit $x_1=31^{-1/5}\Rightarrow x_1^{-5}=31\Rightarrow w_1=32=2^5\Rightarrow w_1^{4/5}=2^4=16$.
   6. Upper limit $x_2=242^{-1/5}\Rightarrow x_2^{-5}=242\Rightarrow w_2=243=3^5\Rightarrow w_2^{4/5}=3^4=81$.
   7. Definite integral $=-\dfrac14(w_2^{4/5}-w_1^{4/5})=-\dfrac14(81-16)=-\dfrac{65}{4}$.

   Common Mistakes

   • Missing the factoring trick and attempting a direct messy substitution.
   • Sign error at the end (dropping the negative sign from $dw=-5x^{-6}dx$), landing on $+65/4$ instead of $-65/4$.

   ✓Final answer

   The correct option is (D) — $\dfrac{-65}{4}$.

   ANSWER: D

4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $\displaystyle\int_0^3 x\left(\sqrt[5]{9 - x^2}\right) dx = k\, 3^{1/k}$, then $k =$ (A) $\dfrac{9}{5}$ (B) $\dfrac{5}{9}$ (C) $\dfrac{5}{12}$ (D) $\dfrac{12}{5}$

   ›Reveal solutionSolution

   The substitution $u=9-x^2$ turns the integral into a simple power integral equal to $\frac{5}{12}\cdot3^{12/5}$, which matches $k\cdot3^{1/k}$ exactly when $k=5/12$.

   Concept and Intuition

   Whenever the integrand contains $x$ times a function of $9-x^2$, the substitution $u=9-x^2$ (so $du=-2x\,dx$) removes the awkward fifth root and reduces the problem to integrating a pure power of $u$.

   Step-by-Step Solution

   1. Let $u=9-x^2 \Rightarrow du=-2x\,dx \Rightarrow x\,dx=-\dfrac{du}{2}$. When $x=0,\ u=9$; when $x=3,\ u=0$.
   2. $\int_0^3 x(9-x^2)^{1/5}dx=\int_{9}^{0}u^{1/5}\left(-\frac{du}{2}\right)=\frac12\int_0^9 u^{1/5}du$
   3. $\frac12\left[\frac{u^{6/5}}{6/5}\right]_0^9=\frac12\cdot\frac56\cdot9^{6/5}=\frac{5}{12}\cdot9^{6/5}$
   4. Since $9=3^2$, $9^{6/5}=3^{12/5}$. So the integral equals $\dfrac{5}{12}\cdot3^{12/5}$.
   5. We need $k\cdot3^{1/k}=\dfrac{5}{12}\cdot3^{12/5}$. Trying $k=\dfrac{5}{12}$: then $1/k=12/5$, giving exactly $\dfrac{5}{12}\cdot3^{12/5}$ — a match.

   Common Mistakes

   • Sign error when flipping the limits after substitution (forgetting the extra minus sign cancels with swapping $9\to0$ to $0\to9$).
   • Not recognizing $9^{6/5}=3^{12/5}$, making it hard to match the $k\cdot 3^{1/k}$ form.

   ✓Final answer

   The correct option is (C) — $\dfrac{5}{12}$.

   ANSWER: C

5. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.$\displaystyle\int_0^2 x^8\left(\dfrac{4}{x^2} - 1\right)^{5/2} dx =$ (A) $\dfrac{2^{15}}{63}$ (B) $\dfrac{2^{16}}{315}$ (C) $\dfrac{2^{16}}{189}$ (D) $\dfrac{2^{10}}{63}$

   ›Reveal solutionSolution

   Simplify the fractional-power term algebraically first (removing the ugly $5/2$ power on a fraction), then a plain $u=4-x^2$ substitution gives $\frac{2^{10}}{63}$.

   Concept and Intuition

   A term like $\left(\frac{4}{x^2}-1\right)^{5/2}$ looks like it needs a trig substitution, but multiplying it out against the accompanying $x^8$ first often collapses the whole thing into a much simpler polynomial-times-power form that a basic $u$-substitution can handle — always simplify algebraically before reaching for a substitution.

   Step-by-Step Solution

   1. Rewrite the bracket: $\dfrac{4}{x^2}-1=\dfrac{4-x^2}{x^2}$, so

   $x^8\left(\frac{4-x^2}{x^2}\right)^{5/2}=x^8\cdot\frac{(4-x^2)^{5/2}}{(x^2)^{5/2}}=x^8\cdot\frac{(4-x^2)^{5/2}}{x^5}=x^3(4-x^2)^{5/2}.$

   2. The integral becomes $\displaystyle I=\int_0^2 x^3(4-x^2)^{5/2}\,dx$.
   3. Substitute $u=4-x^2$, so $du=-2x\,dx$ and $x^2=4-u$. Write $x^3\,dx=x^2(x\,dx)=(4-u)\left(-\frac{du}{2}\right)$.
   4. Limits: $x=0\Rightarrow u=4$; $x=2\Rightarrow u=0$. So

   $I=\int_{u=4}^{0}(4-u)u^{5/2}\left(-\frac12\right)du=\frac12\int_0^4(4-u)u^{5/2}\,du.$

   5. Expand: $\displaystyle I=\frac12\left[4\int_0^4 u^{5/2}du-\int_0^4 u^{7/2}du\right]$.
   6. $\displaystyle\int_0^4 u^{5/2}du=\frac{2}{7}u^{7/2}\Big|_0^4=\frac27\cdot4^{7/2}=\frac27\cdot128=\frac{256}{7}$ (using $4^{7/2}=(2^2)^{7/2}=2^7=128$).
   7. $\displaystyle\int_0^4 u^{7/2}du=\frac{2}{9}u^{9/2}\Big|_0^4=\frac29\cdot4^{9/2}=\frac29\cdot512=\frac{1024}{9}$ (using $4^{9/2}=2^9=512$).
   8. So $\displaystyle I=\frac12\left[4\cdot\frac{256}{7}-\frac{1024}{9}\right]=\frac12\left[\frac{1024}{7}-\frac{1024}{9}\right]=\frac{1024}{2}\left(\frac17-\frac19\right)=512\cdot\frac{2}{63}=\frac{1024}{63}=\frac{2^{10}}{63}.$

   Common Mistakes

   • Attempting a trig substitution ($x=2\sin\theta$) without first simplifying the algebra — it still works but is far more error-prone with the extra $x^8$ power to track.
   • Arithmetic slips converting $4^{7/2}$ and $4^{9/2}$ to powers of $2$ — always write $4=2^2$ first.

   ✓Final answer

   The correct option is (D) — $\dfrac{2^{10}}{63}$.

   ANSWER: D

6. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.$\displaystyle\int_{1/5}^{1/2} \dfrac{\sqrt{x-x^2}}{x^3}\,dx =$ (A) $\dfrac{21}{2}$ (B) $\dfrac{14}{3}$ (C) $\dfrac{7}{3}$ (D) $\dfrac{7}{2}$

   ›Reveal solutionSolution

   This tests factoring $x-x^2=x^2(1/x-1)$ to expose a clean substitution $t=1/x-1$. The definite integral evaluates to $\dfrac{14}{3}$, option (B).

   Concept and Intuition

   The integrand $\dfrac{\sqrt{x-x^2}}{x^3}$ looks like it needs a trig substitution for $\sqrt{x-x^2}$, but factoring out $x^2$ from under the root — valid since $x>0$ on $[1/5,1/2]$ — turns it into $\dfrac{\sqrt{1/x-1}}{x^2}$, a form whose derivative-friendly piece $1/x^2\,dx$ is exactly what appears when differentiating $1/x$. This makes $t=1/x-1$ the natural substitution.

   Step-by-Step Solution

   1. Since $x>0$: $\sqrt{x-x^2}=\sqrt{x^2\left(\dfrac1x-1\right)}=x\sqrt{\dfrac1x-1}$.
   2. So the integrand is $\dfrac{x\sqrt{1/x-1}}{x^3}=\dfrac{\sqrt{1/x-1}}{x^2}$.
   3. Let $t=\dfrac1x-1$; then $dt=-\dfrac{1}{x^2}dx$, i.e. $\dfrac{dx}{x^2}=-dt$.
   4. The integral becomes $\int\sqrt t\,(-dt)=-\dfrac23t^{3/2}+c$.
   5. Limits: at $x=\dfrac15$, $t=5-1=4$; at $x=\dfrac12$, $t=2-1=1$.
   6. Evaluate: $\left[-\dfrac23t^{3/2}\right]_{t=4}^{t=1}=-\dfrac23(1)^{3/2}-\left(-\dfrac23(4)^{3/2}\right)=-\dfrac23+\dfrac23(8)=-\dfrac23+\dfrac{16}{3}=\dfrac{14}{3}$.

   Common Mistakes

   • Forgetting the sign flip from $dt=-dx/x^2$, which would give the negative of the correct answer.
   • Mixing up which limit of $x$ corresponds to which limit of $t$ (note $t$ decreases as $x$ increases here).

   ✓Final answer

   The correct option is (B) — $\dfrac{14}{3}$.

   ANSWER: B

7. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.$\displaystyle\int_1^3 x\sqrt[n]{x^2-1}\,dx = 6$ then $n =$ (A) $2$ (B) $3$ (C) $4$ (D) $5$

   ›Reveal solutionSolution

   The substitution $u=x^2-1$ converts the integral into a power-rule form; testing $n=3$ gives exactly $6$, matching the given value.

   Concept and Intuition

   $x\,dx$ is (up to a constant) exactly the differential of $x^2-1$, so this integral is a textbook power-substitution problem, giving a clean closed form in terms of $n$ that we can then solve (or test) against the given value $6$.

   Step-by-Step Solution

   1. Let $u=x^2-1 \Rightarrow du=2x\,dx \Rightarrow x\,dx=\dfrac{du}{2}$. Limits: $x=1\Rightarrow u=0$; $x=3\Rightarrow u=8$.
   2. $\displaystyle\int_1^3 x(x^2-1)^{1/n}\,dx = \frac12\int_0^8 u^{1/n}\,du = \frac12\left[\frac{u^{1/n+1}}{1/n+1}\right]_0^8 = \frac12\cdot\frac{n}{n+1}\cdot8^{(n+1)/n}$.
   3. Set this equal to $6$ and test integer options. For $n=3$: $8^{4/3} = (8^{1/3})^4 = 2^4=16$. Then $\dfrac12\cdot\dfrac{3}{4}\cdot16 = \dfrac{3}{8}\cdot16=6$ — exact match.

   Common Mistakes

   • Forgetting the $\frac12$ factor from $x\,dx=du/2$.
   • Computing $8^{(n+1)/n}$ incorrectly (mixing up the exponent as $n/(n+1)$ instead of $(n+1)/n$).

   ✓Final answer

   The correct option is (B) — $3$.

   ANSWER: B

8. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.$\displaystyle\int_{0}^{\pi^2/4} \left(2\sin\sqrt{x} + \sqrt{x}\cos\sqrt{x}\right) dx =$ (A) $\pi/2$ (B) $\pi$ (C) $\pi^2/2$ (D) $\pi^2$

   ›Reveal solutionSolution

   Substituting $u=\sqrt x$ turns the integrand into an exact derivative of $u^2\sin u$, giving $\pi^2/2$.

   Concept and Intuition

   When an integrand mixes a function of $\sqrt x$ with $\sqrt x$ times another trig function, substituting $u=\sqrt x$ (so $dx=2u\,du$) often reveals it's the derivative of a simple product — here $u^2\sin u$.

   Step-by-Step Solution

   1. Let $u=\sqrt x \Rightarrow x=u^2,\ dx=2u\,du$. When $x=0,u=0$; when $x=\pi^2/4,\ u=\pi/2$.
   2. The integrand $2\sin\sqrt x + \sqrt x\cos\sqrt x$ becomes $2\sin u + u\cos u$, and with $dx=2u\,du$: $(2\sin u+u\cos u)(2u\,du) = (4u\sin u+2u^2\cos u)\,du$.
   3. Notice $\dfrac{d}{du}(u^2\sin u) = 2u\sin u + u^2\cos u$, so $2\cdot\dfrac{d}{du}(u^2\sin u) = 4u\sin u+2u^2\cos u$ — exactly our integrand.
   4. So the integral is $2\left[u^2\sin u\right]_0^{\pi/2} = 2\left[(\pi/2)^2\sin(\pi/2) - 0\right] = 2\cdot\dfrac{\pi^2}{4}\cdot1 = \dfrac{\pi^2}{2}$.

   Common Mistakes

   • Trying integration by parts directly in $x$ with $\sqrt x$ terms — much messier than the $u=\sqrt x$ substitution.
   • Forgetting the factor of $2u$ from $dx=2u\,du$.

   ✓Final answer

   The correct option is (C) — $\pi^2/2$.

   ANSWER: C

9. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{1}{1+\sqrt{\tan x}} \, dx =$ (A) $0$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{4}$

   ›Reveal solutionSolution

   The classic King's-rule trick ($x\to a-x$) makes the integrand add to $1$; the integral evaluates to $\dfrac\pi4$ — (D).

   Concept and Intuition

   For $\displaystyle\int_0^a f(x)dx$, substituting $x\to a-x$ gives an equal integral $\displaystyle\int_0^a f(a-x)dx$. When $f(x)+f(a-x)$ simplifies to a constant, adding the two versions of the integral collapses everything to a trivial computation.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^{\pi/2}\dfrac{dx}{1+\sqrt{\tan x}}$.
   2. By the property $\displaystyle\int_0^a f(x)dx=\int_0^a f(a-x)dx$: $I=\displaystyle\int_0^{\pi/2}\dfrac{dx}{1+\sqrt{\tan(\pi/2-x)}}=\int_0^{\pi/2}\dfrac{dx}{1+\sqrt{\cot x}}$.
   3. Simplify: $\dfrac1{1+\sqrt{\cot x}}=\dfrac1{1+1/\sqrt{\tan x}}=\dfrac{\sqrt{\tan x}}{\sqrt{\tan x}+1}$.
   4. Adding the two expressions for $I$: $2I=\displaystyle\int_0^{\pi/2}\left[\dfrac1{1+\sqrt{\tan x}}+\dfrac{\sqrt{\tan x}}{1+\sqrt{\tan x}}\right]dx=\int_0^{\pi/2}\dfrac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}}dx=\int_0^{\pi/2}1\,dx=\dfrac\pi2$.
   5. So $I=\dfrac\pi4$.

   Common Mistakes

   • Attempting to directly integrate $1/(1+\sqrt{\tan x})$ via substitution (possible but far messier than the symmetry trick).
   • Sign/reciprocal slip converting $\cot x$ to $1/\tan x$ inside the nested square root.

   ✓Final answer

   The correct option is (D) — $\dfrac\pi4$.

   ANSWER: D

10. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.If $5f(x)+3f\left(\dfrac{1}{x}\right)=2-\dfrac{1}{x}$, $x\ne0$, then $\displaystyle\int_1^2 f\left(\dfrac{1}{x}\right)dx=$ (A) $\dfrac{6\log2-7}{32}$ (B) $\dfrac{6\log2-17}{32}$ (C) $\dfrac{6\log2-1}{32}$ (D) $\dfrac{6\log2-7}{16}$

    ›Reveal solutionSolution

    Replacing $x$ by $1/x$ in the functional equation gives a second equation; solving the linear system for $f(1/x)$ and integrating over $[1,2]$ gives $\dfrac{6\log2-7}{32}$.

    Concept and Intuition

    A functional equation relating $f(x)$ and $f(1/x)$ is solved by generating a second equation (via the substitution $x\to1/x$) and treating $f(x), f(1/x)$ as two unknowns in a linear system.

    Step-by-Step Solution

    1. Given: $5f(x)+3f(1/x)=2-1/x$ … (1)
    2. Replace $x\to1/x$: $5f(1/x)+3f(x)=2-x$ … (2)
    3. Compute $5\times(1)-3\times(2)$: wait — instead eliminate $f(x)$: multiply (1) by $3$ and (2) by $5$: $15f(x)+9f(1/x)=6-3/x$ and $15f(x)+25f(1/x)=10-5x$.
    4. Subtract: $16f(1/x)=4-5x+3/x\Rightarrow f(1/x)=\dfrac{4-5x+3/x}{16}$.
    5. $\displaystyle\int_1^2 f(1/x)\,dx=\dfrac1{16}\int_1^2\left(4-5x+\dfrac3x\right)dx$.
    6. $\int_1^2 4\,dx=4$; $\int_1^2(-5x)\,dx=-5\left[\dfrac{x^2}2\right]_1^2=-5(2-0.5)=-7.5$; $\int_1^2\dfrac3x\,dx=3\log2$.
    7. Sum $=4-7.5+3\log2=3\log2-3.5$; dividing by $16$: $\dfrac{3\log2-3.5}{16}=\dfrac{6\log2-7}{32}$.

    Common Mistakes

    • Solving for $f(x)$ instead of $f(1/x)$ (the question needs $f(1/x)$, integrated in $x$).
    • Arithmetic slip converting $-3.5/16$ to a fraction with denominator $32$.

    ✓Final answer

    The correct option is (A) — $\dfrac{6\log2-7}{32}$.

    ANSWER: A

11. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
    Q.If $\int_1^4 x\sqrt{x^2-1}\, dx = \alpha(k)^\beta$, then $\alpha\beta =$ ________ (A) $\frac{9}{2}$ (B) $\frac{1}{2}$ (C) $\frac{1}{3}$ (D) $\frac{3}{2}$

    ›Reveal solutionSolution

    A simple $u$-substitution evaluates the definite integral to $\frac13(15)^{3/2}$; matching to $\alpha(k)^\beta$ gives $\alpha\beta=\frac12$. Answer: $\frac12$.

    Concept and Intuition

    $x\,dx$ next to $\sqrt{x^2-1}$ is the classic signal for the substitution $u=x^2-1$, turning the integral into a simple power-rule integral in $u$.

    Step-by-Step Solution

    1. Let $u=x^2-1\Rightarrow du=2x\,dx$. When $x=1,u=0$; when $x=4,u=15$.
    2. $\displaystyle\int_1^4 x\sqrt{x^2-1}\,dx=\frac12\int_0^{15}\sqrt u\,du = \frac12\left[\frac23 u^{3/2}\right]_0^{15}=\frac13(15)^{3/2}$.
    3. Matching to $\alpha(k)^\beta$: $\alpha=\dfrac13,\ k=15,\ \beta=\dfrac32$.
    4. $\alpha\beta = \dfrac13\times\dfrac32=\dfrac12$.

    Common Mistakes

    • Forgetting the factor of $\frac12$ that comes from $du=2x\,dx$.

    ✓Final answer

    The correct option is (B) — $\frac12$.

    ANSWER: B

12. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.Let $T > 0$ be a fixed number. $f: \mathbb{R} \to \mathbb{R}$ is a continuous function such that $f(x+T) = f(x)$ $\forall x \in \mathbb{R}$. If $I = \displaystyle\int_0^T f(x)\,dx$, then $\displaystyle\int_0^{5T} f(2x)\,dx =$ (A) $10\,I$ (B) $\dfrac{5}{2}\,I$ (C) $5\,I$ (D) $2\,I$

    ›Reveal solutionSolution

    A substitution $u=2x$ turns the integral into one over $10$ full periods of $f$, giving $5I$.

    Concept and Intuition

    If $f$ is periodic with period $T$, the integral of $f$ over any interval of length $nT$ (a whole number of periods) equals $n$ times the integral over one period. Stretching the variable via $u=2x$ effectively doubles the length of the interval in $u$-space.

    Step-by-Step Solution

    1. Let $u=2x \Rightarrow du = 2\,dx \Rightarrow dx = \dfrac{du}{2}$. When $x=0$, $u=0$; when $x=5T$, $u=10T$.
    2. $\displaystyle\int_0^{5T}f(2x)\,dx = \int_0^{10T}f(u)\cdot\frac{du}{2} = \frac12\int_0^{10T}f(u)\,du$.
    3. Since $f(u+T)=f(u)$ for all $u$, the interval $[0,10T]$ is exactly $10$ periods, so $\int_0^{10T}f(u)\,du = 10\int_0^Tf(u)\,du = 10I$.
    4. So the value is $\dfrac12\cdot10I = 5I$.

    Common Mistakes

    • Forgetting to rescale the upper limit of integration when substituting $u=2x$ (using $5T$ instead of $10T$).
    • Missing the factor of $\frac12$ that comes from $dx=du/2$.

    ✓Final answer

    The correct option is (C) — $5\,I$.

    ANSWER: C