Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{2}x\sqrt{2-x}\,dx$

Showing the 12 most recent of 48 on this concept.

1. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi} \dfrac{x\tan x}{\sec x + \tan x}\, dx =$ (A) $\dfrac{\pi(\pi-2)}{2}$ (B) $\dfrac{\pi+2}{2}$ (C) $\dfrac{\pi(\pi+2)}{2}$ (D) $\dfrac{\pi-2}{2}$

   ›Reveal solutionSolution

   The integrand times $x$ over $[0,\pi]$ is handled with the symmetry trick $\int_0^a xf(x)dx=\frac{a}{2}\int_0^af(x)dx$ when $f(a-x)=f(x)$.

   Concept and Intuition

   $\tan x/(\sec x+\tan x)$ simplifies to $\sin x/(1+\sin x)$, a function symmetric under $x\to\pi-x$. This symmetry lets us replace the troublesome factor of $x$ with a constant $\pi/2$, reducing the problem to a standard definite integral.

   Step-by-Step Solution

   1. Simplify: $\dfrac{\tan x}{\sec x+\tan x}=\dfrac{\sin x/\cos x}{(1+\sin x)/\cos x}=\dfrac{\sin x}{1+\sin x}=f(x)$.
   2. Check symmetry: $f(\pi-x)=\dfrac{\sin(\pi-x)}{1+\sin(\pi-x)}=\dfrac{\sin x}{1+\sin x}=f(x)$. ✓
   3. By the King's rule, $I=\displaystyle\int_0^\pi xf(x)dx=\dfrac{\pi}{2}\int_0^\pi f(x)dx$.
   4. Compute $J=\displaystyle\int_0^\pi\dfrac{\sin x}{1+\sin x}dx=\int_0^\pi\left[1-\dfrac{1}{1+\sin x}\right]dx=\pi-\int_0^\pi\dfrac{dx}{1+\sin x}$. …

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2. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.$\displaystyle\int_{-a}^{a} f(x)\,dx - \int_{0}^{a} f(-x)\,dx =$ (A) $\displaystyle\int_{-a}^{a} f(a-x)\,dx$ (B) $\displaystyle\int_{-a}^{a} f(x) + f(a-x)\,dx$ (C) $\displaystyle\int_{0}^{a} f(x) + f(a-x)\,dx$ (D) $\displaystyle\int_{0}^{a} f(a-x)\,dx$

   ›Reveal solutionSolution

   The expression simplifies to $\int_0^a f(x)dx$, which by the standard reflection property equals $\int_0^a f(a-x)dx$.

   Concept and Intuition

   Split the symmetric integral at the origin and use the substitution $x\to -x$ on the negative half to relate it to $\int_0^a f(-x)dx$ — this is exactly the term being subtracted, so it cancels, leaving a plain $\int_0^a f(x)dx$. Then apply the classic property $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ (true for any integrable $f$, by substituting $x\to a-x$) to match it to the given answer choices.

   Step-by-Step Solution

   1. Split: $\displaystyle\int_{-a}^a f(x)dx = \int_{-a}^0 f(x)dx + \int_0^a f(x)dx$.
   2. In $\int_{-a}^0 f(x)dx$, substitute $x=-t$, $dx=-dt$; limits $x=-a\to t=a$, $x=0\to t=0$: $\displaystyle\int_{-a}^0 f(x)dx = \int_a^0 f(-t)(-dt) = \int_0^a f(-t)dt = \int_0^a f(-x)dx$.
   3. So $\displaystyle\int_{-a}^a f(x)dx = \int_0^a f(-x)dx + \int_0^a f(x)dx$.
   4. Subtracting $\int_0^a f(-x)dx$ from both sides (as required by the question) leaves exactly $\displaystyle\int_0^a f(x)dx$. …

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3. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$\displaystyle\int_{-1}^{1} \left(\sqrt{1+x+x^2} - \sqrt{1-x+x^2}\right) dx =$ (A) $2$ (B) $4$ (C) $0$ (D) $8$

   ›Reveal solutionSolution

   The integrand is an odd function of $x$, so its integral over the symmetric interval $[-1,1]$ vanishes.

   Concept and Intuition

   $\int_{-a}^a f(x)dx=0$ whenever $f$ is odd, i.e. $f(-x)=-f(x)$. Recognising this symmetry avoids a painful direct integration of the square roots.

   Step-by-Step Solution

   1. Let $g(x)=\sqrt{1+x+x^2}$. Then $g(-x)=\sqrt{1-x+x^2}$.
   2. The integrand is $f(x)=g(x)-g(-x)$.
   3. Check parity: $f(-x) = g(-x)-g(x) = -\big(g(x)-g(-x)\big) = -f(x)$ — so $f$ is odd.
   4. Hence $\int_{-1}^1 f(x)\,dx = 0$. …

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4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{x\sin 2x}{1 + 4\cos^2 2x}\, dx =$ (A) $\dfrac{\pi}{8}\left(\mathrm{Tan}^{-1} 2\right)$ (B) $\dfrac{\pi}{4}\left(\mathrm{Tan}^{-1} 2\right)$ (C) $\dfrac{\pi}{2}\left(\mathrm{Tan}^{-1} 2\right)$ (D) $\dfrac{\pi}{8}\left(\mathrm{Tan}^{-1} 4\right)$

   ›Reveal solutionSolution

   Use the King's-rule symmetry $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ to eliminate the explicit $x$, then substitute; the result is $\dfrac\pi8\mathrm{Tan}^{-1}2$.

   Concept and Intuition

   Whenever an integrand is $x$ times a function that is itself symmetric (or anti-symmetric) under $x\to a-x$, replacing $x$ by $a-x$ and adding to the original kills the explicit $x$-dependence, leaving a much simpler integral to evaluate.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^{\pi/2}\frac{x\sin2x}{1+4\cos^22x}\,dx$.
   2. Replace $x\to\frac\pi2-x$: since $\sin(\pi-2x)=\sin2x$ and $\cos(\pi-2x)=-\cos2x$ (so $\cos^2$ is unchanged),

   $I=\int_0^{\pi/2}\frac{\left(\frac\pi2-x\right)\sin2x}{1+4\cos^22x}\,dx=\frac\pi2\int_0^{\pi/2}\frac{\sin2x}{1+4\cos^22x}\,dx - I.$

   3. So $2I=\dfrac\pi2 J$ where $J=\displaystyle\int_0^{\pi/2}\frac{\sin2x}{1+4\cos^22x}\,dx$.
   4. Let $w=\cos2x$, $dw=-2\sin2x\,dx$. Limits: $x=0\Rightarrow w=1$; $x=\pi/2\Rightarrow w=-1$.

   $J=\int_{1}^{-1}\frac{-dw/2}{1+4w^2}=\int_{-1}^{1}\frac{dw/2}{1+4w^2}=\int_0^1\frac{dw}{1+4w^2}$

   (using the evenness of the integrand over the symmetric interval). …

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5. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.$\displaystyle\int_{-\pi}^{\pi} \dfrac{2x(1 + \sin x)}{1 + \cos^2 x}\,dx =$ (A) $2\pi$ (B) $\pi^2$ (C) $\pi + 2$ (D) $\pi/2$

   ›Reveal solutionSolution

   Splitting the integrand by parity kills the odd piece immediately; the remaining even piece is handled with the classic "$\int_0^\pi x f(\sin x, \cos x)dx=\frac{\pi}{2}\int_0^\pi f(\sin x,\cos x)dx$"-style symmetry trick, collapsing to a simple arctangent integral.

   Concept and Intuition

   Whenever a definite integral is over a symmetric interval like $[-\pi,\pi]$, always check parity first: odd integrands vanish, and even integrands can be doubled and computed over $[0,\pi]$ only. For integrals of the form $\int_0^\pi x\,g(\sin x,\cos x)\,dx$, the substitution $x\to\pi-x$ often converts the "$x$" factor into "$\pi - x$", letting you solve for the integral in terms of a simpler one without the $x$ multiplier.

   Step-by-Step Solution

   1. Write $\dfrac{2x(1+\sin x)}{1+\cos^2x}=\dfrac{2x}{1+\cos^2x}+\dfrac{2x\sin x}{1+\cos^2x}$.
   2. The first term, $\dfrac{2x}{1+\cos^2x}$, is odd (odd/even = odd), so $\int_{-\pi}^\pi$ of it is $0$.
   3. The second term, $\dfrac{2x\sin x}{1+\cos^2x}$, is even (odd$\times$odd = even, divided by even), so

   $\int_{-\pi}^{\pi}\frac{2x\sin x}{1+\cos^2x}dx = 2\int_0^\pi\frac{2x\sin x}{1+\cos^2x}dx=4\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx.$

   4. Let $J=\displaystyle\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx$. Substituting $x\to\pi-x$ (using $\sin(\pi-x)=\sin x,\ \cos(\pi-x)=-\cos x$, and $\cos^2$ unchanged): $J=\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}dx=\pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx - J.$ …

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6. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.$\displaystyle\int_{-2}^2(4-x^2)^{5/2}\,dx=$ (A) $40\pi$ (B) $20\pi$ (C) $10\pi$ (D) $\dfrac{5\pi}{32}$

   ›Reveal solutionSolution

   A trig substitution $x=2\sin\theta$ reduces this to a standard $\cos^6\theta$ Wallis integral, giving $20\pi$, option (B).

   Concept and Intuition

   Whenever the integrand contains $(a^2-x^2)^{n/2}$ over symmetric limits $[-a,a]$, the substitution $x=a\sin\theta$ removes the square root entirely and turns it into a pure power of cosine, which is evaluated by the Wallis reduction formula.

   Step-by-Step Solution

   1. Let $x=2\sin\theta$, so $dx=2\cos\theta\,d\theta$ and $4-x^2=4\cos^2\theta$. Limits $x=-2\to2$ correspond to $\theta=-\pi/2\to\pi/2$.
   2. $(4-x^2)^{5/2}=(4\cos^2\theta)^{5/2}=4^{5/2}\cos^5\theta=32\cos^5\theta$.
   3. Integral becomes $\displaystyle\int_{-\pi/2}^{\pi/2}32\cos^5\theta\cdot2\cos\theta\,d\theta=64\int_{-\pi/2}^{\pi/2}\cos^6\theta\,d\theta$.
   4. By symmetry, $=64\cdot2\displaystyle\int_0^{\pi/2}\cos^6\theta\,d\theta=128\cdot\frac{5!!}{6!!}\cdot\frac{\pi}{2}=128\cdot\frac{15}{48}\cdot\frac{\pi}{2}$. …

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7. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{\sin(\pi/4+x)+\sin(3\pi/4+x)}{\cos x+\sin x}dx =$ (A) $\dfrac{\pi}{\sqrt{2}}$ (B) $\dfrac{\pi}{2\sqrt{2}}$ (C) $\dfrac{\pi}{3\sqrt{2}}$ (D) $\dfrac{\pi}{4\sqrt{2}}$

   ›Reveal solutionSolution

   Simplify the numerator using angle-addition identities, then use the classic $x\to\pi/2-x$ symmetry trick (King's rule) to evaluate the definite integral without direct antidifferentiation.

   Concept and Intuition

   Symmetric definite integrals over $[0,\pi/2]$ often simplify beautifully using the substitution $x\to\pi/2-x$, which swaps $\sin$ and $\cos$. Adding the original and transformed integrals frequently collapses to something trivial.

   Step-by-Step Solution

   1. Expand: $\sin(\pi/4+x)=\dfrac{\sqrt2}{2}(\cos x+\sin x)$ and $\sin(3\pi/4+x)=\dfrac{\sqrt2}{2}(\cos x-\sin x)$.
   2. Sum $= \dfrac{\sqrt2}{2}\big[(\cos x+\sin x)+(\cos x-\sin x)\big] = \sqrt2\cos x$.
   3. So $I=\displaystyle\int_0^{\pi/2}\dfrac{\sqrt2\cos x}{\cos x+\sin x}dx$.
   4. Substitute $x\to\pi/2-x$: $I=\displaystyle\int_0^{\pi/2}\dfrac{\sqrt2\sin x}{\sin x+\cos x}dx = J$ (same value, by the King's-rule symmetry). …

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8. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.$\int_{\pi/11}^{9\pi/22} \dfrac{dx}{1 + \sqrt{\tan x}} =$ (A) $\pi/4$ (B) $\pi/22$ (C) $\pi/11$ (D) $7\pi/44$

   ›Reveal solutionSolution

   This tests the classic King's-rule substitution $x\to a+b-x$ for definite integrals whose limits sum to $\pi/2$, applied to a $1/(1+\sqrt{\tan x})$ integrand.

   Concept and Intuition

   Whenever the limits of integration sum to a value that makes $\tan x \to \cot x$ under the substitution $x\to a+b-x$ (here $a+b=\pi/2$, so $\tan(\pi/2-x)=\cot x$), adding the original and transformed integrals often produces a simple constant integrand.

   Step-by-Step Solution

   1. Check $a+b$: $\dfrac{\pi}{11}+\dfrac{9\pi}{22} = \dfrac{2\pi}{22}+\dfrac{9\pi}{22}=\dfrac{11\pi}{22}=\dfrac{\pi}{2}$.
   2. Substitute $x\to \dfrac\pi2-x$: $\tan\left(\dfrac\pi2-x\right)=\cot x = \dfrac1{\tan x}$, so the transformed integrand is $\dfrac{1}{1+1/\sqrt{\tan x}}=\dfrac{\sqrt{\tan x}}{1+\sqrt{\tan x}}$.
   3. So $I=\displaystyle\int_a^b\dfrac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx$ as well (same value as the original, by the substitution property). …

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9. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If $I=\displaystyle\int_{-a}^{a}(x^4-2x^2)\,dx$, then I is minimum at $a=$ (A) $2$ (B) $-\sqrt2$ (C) $\sqrt2$ (D) $-2$

   ›Reveal solutionSolution

   Treating $I(a)=\int_{-a}^a(x^4-2x^2)dx$ as a function of $a$ and minimizing it via calculus gives the minimum at $a=\sqrt2$.

   Concept and Intuition

   Here $a$ is not the integration variable but a parameter — the problem wants the value of $a$ that minimizes the resulting function $I(a)$. Since $x^4-2x^2$ is even, the integral simplifies nicely, and then it's a standard single-variable optimization (first derivative zero, second derivative test).

   Step-by-Step Solution

   1. $x^4-2x^2$ is an even function, so $I(a) = \displaystyle\int_{-a}^a(x^4-2x^2)dx = 2\int_0^a(x^4-2x^2)dx$.
   2. $\displaystyle\int_0^a(x^4-2x^2)dx = \frac{a^5}{5}-\frac{2a^3}{3}$, so $I(a) = \frac{2a^5}{5}-\frac{4a^3}{3}$.
   3. $I'(a) = 2a^4-4a^2 = 2a^2(a^2-2)$.
   4. Set $I'(a)=0$: $a=0$ or $a^2=2 \Rightarrow a=\pm\sqrt2$.
   5. $I''(a) = 8a^3-8a = 8a(a^2-1)$.
   6. At $a=\sqrt2$: $I''(\sqrt2)=8\sqrt2(2-1)=8\sqrt2>0$ → local minimum. …

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10. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
    Q.$\displaystyle\int_0^5 x(5 - x)^{20}\, dx =$ (A) $\dfrac{5^{20}}{441}$ (B) $\dfrac{5^{21}}{441}$ (C) $\dfrac{5^{21}}{462}$ (D) $\dfrac{5^{22}}{462}$

    ›Reveal solutionSolution

    The substitution $u=5-x$ turns the integral into two elementary power integrals, giving $\dfrac{5^{22}}{462}$.

    Concept and Intuition

    Whenever an integral has a linear factor times a high power of $(a-x)$, the substitution $u=a-x$ swaps the roles so the high power becomes the simple variable, and the linear factor becomes $(a-u)$ — turning the whole thing into two standard power-rule integrals.

    Step-by-Step Solution

    1. Let $u = 5-x$, so $x=5-u$ and $dx=-du$. Limits: $x=0\Rightarrow u=5$; $x=5\Rightarrow u=0$.
    2. $\displaystyle\int_0^5 x(5-x)^{20}dx = \int_5^0 (5-u)u^{20}(-du) = \int_0^5(5-u)u^{20}\,du$.
    3. Split: $= 5\displaystyle\int_0^5 u^{20}\,du - \int_0^5 u^{21}\,du = 5\left[\dfrac{u^{21}}{21}\right]_0^5 - \left[\dfrac{u^{22}}{22}\right]_0^5 = \dfrac{5\cdot5^{21}}{21} - \dfrac{5^{22}}{22} = \dfrac{5^{22}}{21}-\dfrac{5^{22}}{22}$. …

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

    ›Reveal solutionSolution

    This is the classic "$x\,f(\sin x)$" King's-rule setup; it reduces to $\pi/2$ times a standard arctan integral, giving $\dfrac{\pi^2}4$ — (D).

    Concept and Intuition

    Whenever the integrand has the shape $x\cdot g(\sin x,\cos^2x)$ over $[0,\pi]$ and $g$ is invariant under $x\to\pi-x$ (true here since $\sin(\pi-x)=\sin x$ and $\cos^2(\pi-x)=\cos^2x$), the King's-rule substitution $x\to\pi-x$ lets us replace the "$x$" prefactor with "$\pi-x$", and adding the two forms eliminates the awkward $x$ entirely.

    Step-by-Step Solution

    1. Let $I=\displaystyle\int_0^\pi\dfrac{x\sin x}{1+\cos^2x}dx$ and $f(x)=\dfrac{\sin x}{1+\cos^2x}$.
    2. Check $f(\pi-x)=\dfrac{\sin(\pi-x)}{1+\cos^2(\pi-x)}=\dfrac{\sin x}{1+\cos^2x}=f(x)$ — confirmed invariant.
    3. By King's rule: $I=\displaystyle\int_0^\pi(\pi-x)f(x)\,dx=\pi\int_0^\pi f(x)dx - I$.
    4. So $2I=\pi\displaystyle\int_0^\pi f(x)\,dx=\pi J$, where $J=\displaystyle\int_0^\pi\dfrac{\sin x}{1+\cos^2x}dx$. …

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12. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.$\displaystyle\int_0^{\pi} \dfrac{x\sin x}{1+\cos^2 x}\,dx =$ (A) $\dfrac{\pi^2}{4}$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi^2}{2}$ (D) $\dfrac{\pi}{4}$

    ›Reveal solutionSolution

    This is the classic $\int_0^\pi x f(\sin x,\cos x)\,dx$ trick using the substitution $x\to \pi-x$; the answer is (A).

    Concept and Intuition

    For integrals of the form $\int_0^\pi x\,g(\cos x)\sin x\,dx$, substituting $x\to\pi-x$ (using $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)=-\cos x$, but here $\cos^2$ is unaffected) converts the $x$ factor into $\pi - x$, letting you solve for the integral algebraically in terms of a simpler integral without the $x$ weight.

    Step-by-Step Solution

    1. Let $I=\displaystyle\int_0^\pi\frac{x\sin x}{1+\cos^2 x}dx$.
    2. Substitute $x\to\pi-x$: $\sin(\pi-x)=\sin x$, $\cos^2(\pi-x)=\cos^2 x$, so

    $I=\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}dx=\pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx-I.$

    3. So $2I=\pi\displaystyle\int_0^\pi\frac{\sin x}{1+\cos^2 x}dx$. …

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