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

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.$\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} \, dx =$ (A) $\frac{1}{2}$ (B) $\frac{3}{2}$ (C) $2$ (D) $1$

   ›Reveal solutionSolution

   This integral is a classic symmetric-invariance problem: using the substitution $x \to 9 - x$ shows the integrand and its complement sum to 1, so the integral over $[3,6]$ is half the interval length, giving $\frac{3}{2}$.

   Concept & Intuition

   When you see an integral of the form $\int_a^b \frac{f(x)}{f(x) + f(a+b-x)} \, dx$, there’s a beautiful trick: the integrand and its “mirror image” add to 1. Here $a=3$, $b=6$, so $a+b=9$. The denominator is $\sqrt{9-x} + \sqrt{x}$, and the numerator is $\sqrt{x}$. If we replace $x$ by $9-x$, the numerator becomes $\sqrt{9-x}$ and the denominator stays the same (just swapped order). So the original integrand $I(x)$ and $I(9-x)$ sum to 1. Integrating over a symmetric interval around the midpoint $x=4.5$ then gives half the length of the interval.

   Step-by-step solution

   1. Define the integral Let

   $I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx.$

   2. Apply the substitution $x \to 9 - u$ Set $u = 9 - x$. Then $dx = -du$, and when $x=3$, $u=6$; when $x=6$, $u=3$. So

   $I = \int_{6}^{3} \frac{\sqrt{9-u}}{\sqrt{u} + \sqrt{9-u}} \, (-du) = \int_{3}^{6} \frac{\sqrt{9-u}}{\sqrt{u} + \sqrt{9-u}} \, du.$

   Since $u$ is a dummy variable, rename it $x$:

   $I = \int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx.$

   3. Add the two expressions for $I$ We now have two representations:

   $I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx \quad \text{and} \quad I = \int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx.$

   Adding them:

   $$2I = \int_{3}^{6} \left( \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} + \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \right) dx. …$$

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2. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.$\int_{0}^{2a} f(x) dx =$ (A) $2 \int_{0}^{a} f(x) dx$ (B) $\int_{0}^{a} (f(x) + f(x+a)) dx$ (C) $0$ (D) $\int_{0}^{2a} f(2a + x) dx$

   ›Reveal solutionSolution

   The key idea is to split the integral at $x = a$ and use a substitution $x = a + t$ to relate the two halves. The correct answer is $\int_{0}^{2a} f(x) dx = \int_{0}^{a} (f(x) + f(x+a)) dx$, which is option (B).

   The question asks for a general property of definite integrals over an interval of length $2a$. This is a standard result often used in problems involving periodic or symmetric functions, but it holds for any integrable function $f$. The trick is to break the interval into two equal halves and then shift the second half back to start at $0$.

   1. Split the integral at the midpoint. Write the integral from $0$ to $2a$ as the sum of two integrals:

   $\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{a}^{2a} f(x) dx.$

   This is always valid because the limits are contiguous.

   2. Change variable in the second integral. In $\int_{a}^{2a} f(x) dx$, let $x = a + t$. Then $dx = dt$, and when $x = a$, $t = 0$; when $x = 2a$, $t = a$. So

   $\int_{a}^{2a} f(x) dx = \int_{0}^{a} f(a + t) dt.$

   Since the variable of integration is dummy, we can rename $t$ back to $x$:

   $\int_{a}^{2a} f(x) dx = \int_{0}^{a} f(x + a) dx.$

   3. Combine the two pieces. Putting it together:

   $\int_{0}^{2a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} f(x + a) dx = \int_{0}^{a} \bigl( f(x) + f(x + a) \bigr) dx.$

   This matches option (B) exactly. …

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3. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Let $f:[0,1] \to \mathbb{R}$ be a function defined as $f(x) + f(1-x) = 1$. Then $\int_{0}^{1} f(x) dx =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   The functional equation $f(x)+f(1-x)=1$ forces the average value of $f$ over $[0,1]$ to be $\frac12$, so the integral is $\frac12$. The correct option is (C).

   Concept & Intuition

   The given condition $f(x)+f(1-x)=1$ is a symmetry relation: the value at $x$ and the value at its mirror point $1-x$ always sum to $1$. This means the graph of $f$ is symmetric about the point $\left(\frac12,\frac12\right)$. If you average $f$ over the whole interval, the contributions from $x$ and $1-x$ together always give $1$, so the overall average must be $\frac12$. The integral is just the average value times the length of the interval.

   ---

   1. Set up the integral and use the substitution $x \to 1-x$. Let $I = \int_0^1 f(x)\,dx$. Substitute $u = 1-x$, so $du = -dx$ and when $x=0$, $u=1$; when $x=1$, $u=0$. Then

   $I = \int_0^1 f(x)\,dx = \int_1^0 f(1-u)\,(-du) = \int_0^1 f(1-u)\,du.$

   Renaming the dummy variable back to $x$, we have

   $I = \int_0^1 f(1-x)\,dx.$

   2. Add the two expressions for $I$. We now have two representations:

   $I = \int_0^1 f(x)\,dx \quad\text{and}\quad I = \int_0^1 f(1-x)\,dx.$

   Adding them gives

   $2I = \int_0^1 \bigl[f(x) + f(1-x)\bigr]\,dx.$

   3. Use the given functional equation. The condition $f(x)+f(1-x)=1$ holds for every $x\in[0,1]$. Therefore $$2I = \int_0^1 1\,dx = \bigl[x\bigr]_{0}^{1} = 1. …$$

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4. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If $\int_{-a}^{a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} g(x) dx$ then $g(x) =$ (A) $-f(x)$ (B) $f(x)$ (C) $f(-x)$ (D) $f(x) + f(-x)$

   ›Reveal solutionSolution

   The property of definite integrals over symmetric limits splits into two parts: one from $0$ to $a$ of $f(x)$, and another from $0$ to $a$ of $f(-x)$. So $g(x) = f(-x)$, which is option (C).

   The key idea here is how an integral over a symmetric interval $[-a, a]$ can be broken into two integrals from $0$ to $a$ — one using the original function and the other using a reflected version. This is a standard trick in definite integration, especially when dealing with even and odd functions, but it works for any function.

   Let’s see why this works.

   1. Split the original integral at zero. For any function $f$, we can write:

   $\int_{-a}^{a} f(x) \, dx = \int_{-a}^{0} f(x) \, dx + \int_{0}^{a} f(x) \, dx$

   2. Change variable in the first integral. Let $x = -t$. Then $dx = -dt$, and when $x = -a$, $t = a$; when $x = 0$, $t = 0$. So:

   $\int_{-a}^{0} f(x) \, dx = \int_{a}^{0} f(-t) \, (-dt) = \int_{0}^{a} f(-t) \, dt$

   The limits flip and the minus signs cancel. Since the variable of integration is dummy, we can rename $t$ back to $x$:

   $\int_{-a}^{0} f(x) \, dx = \int_{0}^{a} f(-x) \, dx$

   3. Combine the two pieces. Substituting back:

   $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} f(-x) \, dx + \int_{0}^{a} f(x) \, dx$

   4. Compare with the given form. The problem states: $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} g(x) \, dx$ …

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5. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.

   [!FORMULA] $\int_{0}^{\pi} x f(\sin x) dx =$

   (A) $2\pi \int_{0}^{\pi/4} f(\sin x) dx$ (B) $\pi \int_{0}^{\pi/4} f(\sin x) dx$ (C) $2\pi \int_{0}^{\pi/2} f(\sin x) dx$ (D) $\pi \int_{0}^{\pi/2} f(\sin x) dx$

   ›Reveal solutionSolution

   Use the property $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ with $a=\pi$ to rewrite the integral, then add the two forms. The result is $\pi \int_0^{\pi} f(\sin x)\,dx$, which simplifies to $\pi \int_0^{\pi/2} f(\sin x)\,dx$ because $f(\sin x)$ is symmetric about $\pi/2$. The correct option is (D).

   The key idea here is a classic trick for integrals of the form $\int_0^a x\,g(x)\,dx$: replace $x$ by $a-x$ and add the two expressions. This often cancels the $x$ factor and leaves a simpler integral.

   Let $I = \int_{0}^{\pi} x \, f(\sin x) \, dx$. The function $f(\sin x)$ depends on $x$ only through $\sin x$, which has the property $\sin(\pi - x) = \sin x$. That symmetry is what we will exploit.

   1. Apply the substitution $x \to \pi - x$. Let $t = \pi - x$. Then $dx = -dt$, and when $x=0$, $t=\pi$; when $x=\pi$, $t=0$. So

   $I = \int_{0}^{\pi} x \, f(\sin x) \, dx = \int_{\pi}^{0} (\pi - t) \, f(\sin(\pi - t)) \, (-dt) = \int_{0}^{\pi} (\pi - t) \, f(\sin t) \, dt.$

   Since the dummy variable doesn’t matter, rename $t$ back to $x$:

   $I = \int_{0}^{\pi} (\pi - x) \, f(\sin x) \, dx.$

   2. Add the two expressions for $I$. We now have two forms:

   $I = \int_{0}^{\pi} x \, f(\sin x) \, dx \quad \text{and} \quad I = \int_{0}^{\pi} (\pi - x) \, f(\sin x) \, dx.$

   Adding them:

   $2I = \int_{0}^{\pi} \bigl[ x + (\pi - x) \bigr] f(\sin x) \, dx = \int_{0}^{\pi} \pi \, f(\sin x) \, dx.$

   Hence

   $I = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx.$

   3. Simplify the limits using symmetry. …

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6. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.$\int_{0}^{2} x^3 (2-x)^4 \, dx =$ (A) $\dfrac{128}{105}$ (B) $\dfrac{16}{35}$ (C) $\dfrac{256}{105}$ (D) $\dfrac{32}{35}$

   ›Reveal solutionSolution

   Substitute $u=2-x$ and expand; the integral evaluates to $\dfrac{32}{35}$.

   Setup. Let $u = 2-x$, so $x = 2-u$ and $dx = -du$. The limits map $x:0\to2$ into $u:2\to0$:

   $\int_{0}^{2} x^3(2-x)^4\,dx = \int_{0}^{2} (2-u)^3\,u^4\,du.$

   Expand $(2-u)^3 = 8 - 12u + 6u^2 - u^3$, so the integrand becomes

   $8u^4 - 12u^5 + 6u^6 - u^7.$

   Integrate term-by-term from $0$ to $2$:

   $\left[\frac{8}{5}u^5 - 2u^6 + \frac{6}{7}u^7 - \frac{1}{8}u^8\right]_0^2.$

   At $u=2$: $\dfrac{8}{5}(32) - 2(64) + \dfrac{6}{7}(128) - \dfrac{1}{8}(256) = \dfrac{256}{5} - 128 + \dfrac{768}{7} - 32.$ …

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7. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.$\int_{-1}^{1} x |x| dx =$ (A) $1$ (B) $\dfrac{1}{2}$ (C) $0$ (D) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   The integral of an odd function over a symmetric interval is zero. Since $x|x|$ is odd, the definite integral from $-1$ to $1$ is $0$.

   The key idea here is symmetry. When you integrate an odd function over an interval symmetric about zero, the positive and negative contributions cancel perfectly. The function $f(x) = x|x|$ is odd because $f(-x) = (-x)|-x| = -x|x| = -f(x)$. So instead of doing any messy piecewise integration, we can immediately see the result.

   Let’s verify this step by step to be thorough.

   1. Understand the function

      The absolute value makes the function piecewise:

      • For $x \ge 0$, $|x| = x$, so $x|x| = x \cdot x = x^2$.
      • For $x < 0$, $|x| = -x$, so $x|x| = x \cdot (-x) = -x^2$. So $f(x) = \begin{cases} -x^2, & x < 0 \\ x^2, & x \ge 0 \end{cases}$. This confirms it’s odd: the graph for negative $x$ is the mirror image (with opposite sign) of the graph for positive $x$.

   2. Split the integral at the symmetry point

      Since the function changes definition at $x=0$, we write:

   $\int_{-1}^{1} x|x| \, dx = \int_{-1}^{0} (-x^2) \, dx + \int_{0}^{1} x^2 \, dx.$

   3. Evaluate each piece
      • For the left part: $\int_{-1}^{0} -x^2 \, dx = -\left[ \frac{x^3}{3} \right]_{-1}^{0} = -\left(0 - \frac{(-1)^3}{3}\right) = -\left(0 + \frac{1}{3}\right) = -\frac{1}{3}$. …

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8. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.

   [!FORMULA] $\int_{-1}^{1} \frac{\log(1+x)}{1+x^2} dx = \int_{0}^{1} \frac{\log(1+x)}{1+x^2} dx + \int_{0}^{1} f(x) dx \text{ then } f(x) =$

   (A) $\frac{\log(1+x)}{1+x^2}$ (B) $-\frac{\log(1+x)}{1+x^2}$ (C) $\frac{\log(1-x)}{1+x^2}$ (D) $0$

   ›Reveal solutionSolution

   The key idea is to split the integral at $0$ and then use the substitution $x \to -x$ on the negative half to rewrite it as an integral from $0$ to $1$; the function $f(x)$ turns out to be $\frac{\log(1-x)}{1+x^2}$, which is option (C).

   The problem gives you a split of the original integral from $-1$ to $1$ into two parts: one from $-1$ to $0$ and one from $0$ to $1$. The second part is already written as $\int_0^1 \frac{\log(1+x)}{1+x^2} dx$. The first part, $\int_{-1}^0 \frac{\log(1+x)}{1+x^2} dx$, is what needs to be transformed into $\int_0^1 f(x) dx$. So we need to find $f(x)$ such that

   $\int_{-1}^0 \frac{\log(1+x)}{1+x^2} dx = \int_0^1 f(x) dx.$

   The natural way to convert an integral over a negative interval to one over a positive interval is a change of variable that flips the limits. Let’s work through it.

   1. Set up the substitution. On the interval $[-1, 0]$, let $x = -t$. Then when $x = -1$, $t = 1$; when $x = 0$, $t = 0$. Also $dx = -dt$. The integral becomes

   $\int_{-1}^0 \frac{\log(1+x)}{1+x^2} dx = \int_{1}^{0} \frac{\log(1 - t)}{1 + t^2} (-dt) = \int_{0}^{1} \frac{\log(1 - t)}{1 + t^2} dt.$

   The minus sign from $dx = -dt$ flips the limits back to $0$ to $1$, and $x^2 = t^2$ so the denominator is unchanged.

   2. Identify $f(x)$. The variable of integration is a dummy, so rename $t$ back to $x$. We have

   $\int_{-1}^0 \frac{\log(1+x)}{1+x^2} dx = \int_0^1 \frac{\log(1 - x)}{1 + x^2} dx.$

   Therefore, the function $f(x)$ that makes the original equation hold is

   $f(x) = \frac{\log(1 - x)}{1 + x^2}.$ …

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9. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.$\int_{-\pi/15}^{\pi/15} \frac{\cos 5x}{1 + e^{5x}} \, dx =$ (A) $\frac{1}{5}$ (B) $\frac{\sqrt{3}}{10}$ (C) $\frac{1}{15}$ (D) $\frac{1}{10}$

   ›Reveal solutionSolution

   The symmetry trick $\int_{-a}^{a}\frac{f(x)}{1+e^{cx}}dx=\int_0^a f(x)\,dx$ (for even $f$) reduces this to $\int_0^{\pi/15}\cos 5x\,dx=\frac{\sqrt3}{10}$.

   Use the king-property symmetry. Let

   $I=\int_{-\pi/15}^{\pi/15}\frac{\cos 5x}{1+e^{5x}}\,dx.$

   Replacing $x\to -x$ (limits symmetric) and using $\cos(-5x)=\cos 5x$:

   $I=\int_{-\pi/15}^{\pi/15}\frac{\cos 5x\;e^{5x}}{1+e^{5x}}\,dx.$

   Adding the two forms, since $\dfrac{1}{1+e^{5x}}+\dfrac{e^{5x}}{1+e^{5x}}=1$:

   $2I=\int_{-\pi/15}^{\pi/15}\cos 5x\,dx=2\int_{0}^{\pi/15}\cos 5x\,dx.$

   Evaluate. …

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10. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.$\frac{3}{25} \int_{0}^{25\pi} \sqrt{|\cos x - \cos^3 x|} \, dx =$ (A) $8$ (B) $4$ (C) $1$ (D) $0$

    ›Reveal solutionSolution

    $\sqrt{|\cos x - \cos^3 x|} = \sqrt{|\cos x|}\,|\sin x|$, whose integral over one period $\pi$ is $\tfrac{4}{3}$; over $25\pi$ it is $\tfrac{100}{3}$, and $\tfrac{3}{25}\cdot\tfrac{100}{3} = 4$.

    Simplifying the integrand.

    $\cos x - \cos^3 x = \cos x\,(1-\cos^2 x) = \cos x\,\sin^2 x.$

    Since $\sin^2 x \ge 0$,

    $\sqrt{|\cos x - \cos^3 x|} = \sqrt{|\cos x|\,\sin^2 x} = \sqrt{|\cos x|}\,|\sin x|.$

    Periodicity.

    Both $|\cos x|$ and $|\sin x|$ have period $\pi$, so the integrand has period $\pi$. The interval $[0,25\pi]$ contains exactly $25$ periods.

    Integral over one period $[0,\pi]$.

    On $[0,\pi]$, $\sin x \ge 0$. Split where $\cos x$ changes sign at $\tfrac{\pi}{2}$:

    $\int_0^{\pi}\sqrt{|\cos x|}\,\sin x\,dx = \int_0^{\pi/2}\sqrt{\cos x}\,\sin x\,dx + \int_{\pi/2}^{\pi}\sqrt{-\cos x}\,\sin x\,dx.$ …

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11. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.$\int_{-\pi/8}^{\pi/8} \frac{\sin^4(4x)}{1+e^{4x}} \, dx =$ (A) $\frac{3\pi}{128}$ (B) $\frac{3\pi}{256}$ (C) $\frac{3\pi}{64}$ (D) $\frac{3\pi}{32}$

    ›Reveal solutionSolution

    The $\dfrac{1}{1+e^{4x}}$ symmetry trick reduces the integral to $\displaystyle\int_0^{\pi/8}\sin^4(4x)\,dx=\dfrac{3\pi}{64}$.

    Apply the symmetric-interval identity. For an even function $g$,

    $\int_{-a}^{a}\frac{g(x)}{1+e^{4x}}\,dx=\int_0^{a}g(x)\,dx.$

    This follows from adding $I$ to its $x\to-x$ image: $\dfrac{1}{1+e^{4x}}+\dfrac{1}{1+e^{-4x}}=1$. Here $g(x)=\sin^4(4x)$ is even and $a=\dfrac{\pi}{8}$, so

    $I=\int_0^{\pi/8}\sin^4(4x)\,dx.$

    Evaluate. Substitute $u=4x$, $du=4\,dx$; limits $0\to\pi/2$: …

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12. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.$\int_{0}^{3} \left[ \sin\left(\frac{\pi}{3}x\right) - \cos\left(\frac{\pi}{3}x\right) \right] dx =$ (A) $\dfrac{-6}{\pi}$ (B) $0$ (C) $\dfrac{-3}{\pi}$ (D) $\dfrac{6}{\pi}$

    ›Reveal solutionSolution

    Integrating over one full period-related span $[0,3]$: the $\sin$ term contributes $\tfrac6\pi$ and the $\cos$ term contributes $0$, so the integral is $\tfrac6\pi$ (option D).

    $\int_0^3\left[\sin\!\left(\tfrac{\pi}{3}x\right)-\cos\!\left(\tfrac{\pi}{3}x\right)\right]dx.$

    Sine part:

    $\int_0^3\sin\!\left(\tfrac{\pi}{3}x\right)dx=\left[-\dfrac{3}{\pi}\cos\!\left(\tfrac{\pi}{3}x\right)\right]_0^3=-\dfrac{3}{\pi}\big(\cos\pi-\cos0\big)=-\dfrac3\pi(-1-1)=\dfrac6\pi.$

    Cosine part: …

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