Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{2\pi}\cos^5 x\,dx$

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.$\int_{-2\pi}^{2\pi} \sin^4 x \cos^6 x \, dx =$ (A) $\dfrac{3\pi}{128}$ (B) $\dfrac{9\pi}{32}$ (C) $\dfrac{9\pi}{64}$ (D) $\dfrac{3\pi}{64}$

   ›Reveal solutionSolution

   The integral of an even power of sine and cosine over a full period can be reduced using symmetry and the Beta function; the value is $\frac{3\pi}{64}$, which corresponds to option (D).

   The key insight: the integrand $\sin^4 x \cos^6 x$ is an even function (since both sine and cosine are raised to even powers, the product is symmetric about $x=0$). Also, over $[-2\pi, 2\pi]$, the function repeats its pattern four times (period $\pi$ for the product of even powers). So we can simplify the integral to a multiple of an integral over $[0, \pi/2]$, where the classic Beta-function reduction applies.

   1. Use symmetry and periodicity. The function $f(x) = \sin^4 x \cos^6 x$ has period $\pi$ (because $\sin^2 x$ and $\cos^2 x$ have period $\pi$, and even powers preserve that). Over $[-2\pi, 2\pi]$, which is 4 periods of length $\pi$, we have

   $\int_{-2\pi}^{2\pi} f(x)\,dx = 4 \int_{0}^{\pi} f(x)\,dx.$

   Also, $f(x)$ is even, so $\int_{0}^{\pi} f(x)\,dx = 2\int_{0}^{\pi/2} f(x)\,dx$. Thus

   $\int_{-2\pi}^{2\pi} f(x)\,dx = 8 \int_{0}^{\pi/2} \sin^4 x \cos^6 x \, dx.$

   2. Reduce to a Beta integral. Recall the Beta function identity:

   $\int_{0}^{\pi/2} \sin^{2a-1} x \cos^{2b-1} x \, dx = \frac{1}{2} B(a,b) = \frac{\Gamma(a)\Gamma(b)}{2\,\Gamma(a+b)}.$

   Here we have $\sin^4 x = \sin^{2\cdot 2.5 - 1} x$? Let's match carefully: we want exponents $2a-1 = 4$ and $2b-1 = 6$, so

   $2a-1 = 4 \implies a = \frac{5}{2}, \qquad 2b-1 = 6 \implies b = \frac{7}{2}.$

   Then

   $\int_{0}^{\pi/2} \sin^4 x \cos^6 x \, dx = \frac{1}{2} B\!\left(\frac{5}{2}, \frac{7}{2}\right).$

   3. Evaluate the Beta function using Gamma values.

   $B\!\left(\frac{5}{2}, \frac{7}{2}\right) = \frac{\Gamma(5/2)\,\Gamma(7/2)}{\Gamma(6)}.$

   Recall $\Gamma(n) = (n-1)!$ for integers, and $\Gamma(1/2) = \sqrt{\pi}$.

   $\Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{3\sqrt{\pi}}{4},$

   $\Gamma(7/2) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{15\sqrt{\pi}}{8},$

   $\Gamma(6) = 5! = 120.$

   So

   $$$$

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2. 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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3. 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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4. 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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5. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.$\int\limits_{0}^{\pi} (\sin^3 x + \cos^2 x)^2 dx =$ (A) $\dfrac{15\pi}{16} + \dfrac{8}{15}$ (B) $\dfrac{11\pi}{16} + \dfrac{8}{15}$ (C) $\dfrac{15\pi}{16} + \dfrac{4}{15}$ (D) $\dfrac{11\pi}{16} + \dfrac{4}{15}$

   ›Reveal solutionSolution

   The integral simplifies by expanding the square, using symmetry and reduction formulas; the final value is $\frac{11\pi}{16} + \frac{8}{15}$, which corresponds to option (B).

   We start with

   $I = \int_0^\pi (\sin^3 x + \cos^2 x)^2 \, dx.$

   The key idea: expand the square, then use symmetry and standard trigonometric integrals. The presence of both odd and even powers suggests splitting the interval or using reduction formulas.

   1. Expand the square

   $(\sin^3 x + \cos^2 x)^2 = \sin^6 x + 2\sin^3 x \cos^2 x + \cos^4 x.$

   So

   $I = \int_0^\pi \sin^6 x \, dx + 2\int_0^\pi \sin^3 x \cos^2 x \, dx + \int_0^\pi \cos^4 x \, dx.$

   2. Use symmetry for the middle integral The function $\sin^3 x \cos^2 x$ is odd about $x = \pi/2$? Check: $\sin(\pi - x) = \sin x$, $\cos(\pi - x) = -\cos x$, so $\sin^3(\pi-x)\cos^2(\pi-x) = \sin^3 x \cos^2 x$. Actually it's symmetric, not odd. But better: note that $\sin^3 x \cos^2 x$ is an odd function with respect to $x = \pi/2$? Let's test: replace $x$ by $\pi - x$, the product is unchanged. So it's symmetric. However, we can compute directly using substitution $u = \cos x$:

   $\int_0^\pi \sin^3 x \cos^2 x \, dx = \int_0^\pi \sin x (1-\cos^2 x) \cos^2 x \, dx.$

   Let $u = \cos x$, $du = -\sin x \, dx$, limits: $x=0 \to u=1$, $x=\pi \to u=-1$. Then

   $\int_0^\pi \sin^3 x \cos^2 x \, dx = \int_1^{-1} (1-u^2) u^2 (-du) = \int_{-1}^1 (u^2 - u^4) \, du.$

   The integrand is even, so

   $= 2\int_0^1 (u^2 - u^4) \, du = 2\left[\frac{u^3}{3} - \frac{u^5}{5}\right]_0^1 = 2\left(\frac13 - \frac15\right) = 2\cdot\frac{2}{15} = \frac{4}{15}.$

   Thus

   $2\int_0^\pi \sin^3 x \cos^2 x \, dx = \frac{8}{15}.$

   3. Compute $\int_0^\pi \sin^6 x \, dx$ Use the reduction formula or known result:

   $\int_0^\pi \sin^{2n} x \, dx = \pi \cdot \frac{(2n-1)!!}{(2n)!!}.$

   For $n=3$, $\sin^6 x$ has even power, so

   $\int_0^\pi \sin^6 x \, dx = \pi \cdot \frac{5!!}{6!!} = \pi \cdot \frac{5 \cdot 3 \cdot 1}{6 \cdot 4 \cdot 2} = \pi \cdot \frac{15}{48} = \frac{5\pi}{16}.$

   (Check: $5!! = 15$, $6!! = 48$, yes.)

   4. Compute $\int_0^\pi \cos^4 x \, dx$ Since $\cos^4 x$ is symmetric about $\pi/2$, we can also use $\int_0^\pi \cos^4 x \, dx = 2\int_0^{\pi/2} \cos^4 x \, dx$. Use the reduction formula for $\int_0^{\pi/2} \cos^{2n} x \, dx = \frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2}$. …

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6. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x (\sin x + \cos x) dx =$ (A) $\dfrac{2}{3}$ (B) $\dfrac{3}{10}$ (C) $\dfrac{4}{15}$ (D) $\dfrac{5}{18}$

   ›Reveal solutionSolution

   The integrand is an odd function over a symmetric interval, so the integral evaluates to zero; none of the given positive options match, but the correct answer is 0, which is not listed — the intended answer is (C) only if the problem had a misprint, but strictly the integral is zero.

   The key insight is symmetry. When integrating over $[-\pi/2, \pi/2]$, check if the function is odd or even. An odd function integrated over a symmetric interval always gives zero. Here, $\sin^2 x \cos^2 x$ is even, but $(\sin x + \cos x)$ is a sum of an odd and an even part. The product of an even function with an odd function is odd, and that part integrates to zero. The even part (from $\cos x$) also integrates to zero because of the specific powers? Let’s check carefully.

   1. Separate the integrand:

   $\sin^2 x \cos^2 x (\sin x + \cos x) = \sin^2 x \cos^2 x \sin x + \sin^2 x \cos^2 x \cos x.$

   2. Analyze parity:

      • $\sin^2 x \cos^2 x$ is even because $\sin^2 x$ and $\cos^2 x$ are both even.
      • $\sin x$ is odd, so $\sin^2 x \cos^2 x \cdot \sin x$ is odd.
      • $\cos x$ is even, so $\sin^2 x \cos^2 x \cdot \cos x$ is even.

   3. Integrate the odd part:

      For any odd function $f(x)$, $\int_{-a}^{a} f(x)\,dx = 0$.

      Thus,

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

   4. Integrate the even part: The even part is $\sin^2 x \cos^3 x$. Over a symmetric interval, we can double the integral from $0$ to $\pi/2$: $\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^3 x \, dx = 2 \int_{0}^{\pi/2} \sin^2 x \cos^3 x \, dx.$ …

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7. 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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8. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If $m,\, l,\, r,\, s,\, n$ are integers such that, $9 > m > l > s > n > r > 2$ and

   [!FORMULA] $\int_{-\pi}^{\pi} \sin^r x \cos^s x\, dx = 4 \left| \int_{-\pi}^{\pi} \sin^l x \cos^r x\, dx \right|, \quad \int_{0}^{\frac{7\pi}{2}} \sin^r x \cos^s x\, dx = 4 \left| \int_{0}^{\pi} \sin^l x \cos^r x\, dx \right| \text{ and}$

   [!FORMULA] $\int_{-\frac{\pi}{2}}^{\frac{7\pi}{2}} \sin^l x \cos^m x\, dx = 0,$

   then (A) $(s-2)(l-2) = mr$ (B) $(s-2)(l+2) = rm + 5$ (C) $(s-2)(s+2) = ln - 3$ (D) $(l-2)(l+2) = ms - 5$

   ›Reveal solutionSolution

   The chain of inequalities leaves only six integer sets, and the symmetry conditions on the trigonometric integrals fix the parities. Checking the four printed relations, only $(s-2)(s+2)=log-3$ survives — option (C).

   The concept first

   Two symmetry facts do all the work.

   1. An odd power of sine kills a symmetric integral. $\sin x$ is odd, $\cos x$ is even, so $\sin^{p}x\cos^{q}x$ is an odd function when $p$ is odd, and therefore $\int_{-\pi}^{\pi}\sin^{p}x\cos^{q}x\,dx=0$.
   2. An odd power of cosine kills the $[0,\pi]$ integral. Under $x\mapsto\pi-x$, $\sin x$ is unchanged but $\cos x\mapsto-\cos x$; so $\int_{0}^{\pi}\sin^{p}x\cos^{q}x\,dx=0$ when $q$ is odd.

   Combining: on $[-\pi,\pi]$ the integral is non-zero only if both exponents are even, and then

   $\int_{-\pi}^{\pi}\sin^{p}x\cos^{q}x\,dx=2\int_{0}^{\pi}\sin^{p}x\cos^{q}x\,dx=4\int_{0}^{\pi/2}\sin^{p}x\cos^{q}x\,dx .$

   That is exactly the shape of the given equations — the factor $4$ is a symmetry statement, not an accident. The condition that the third integral (taken over a whole number of periods) is $0$ forces one of $l,m$ to be odd.

   Step 1 — List the candidate sets

   All five integers lie strictly between $2$ and $9$, i.e. in $\{3,4,5,6,7,8\}$, and are strictly increasing in the order $r<n<s<l<m$. So we just drop one member of that six-element set:

   $(r,n,s,l,m)\in\{(4,5,6,7,8),(3,5,6,7,8),(3,4,6,7,8),(3,4,5,7,8),(3,4,5,6,8),(3,4,5,6,7)\}.$

   Step 2 — Apply the parity conditions …

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9. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.$\int_{-2}^{2} x^4 (4-x^2)^7 dx =$ (A) $4\pi$ (B) $\dfrac{\pi}{16}$ (C) $28\pi$ (D) $\dfrac{3\pi}{128}$

   ›Reveal solutionSolution

   With the even integrand and the substitution $x=2\sin\theta$, the integral reduces to a Beta function. For the value to be a multiple of $\pi$ the intended power is $(4-x^2)^{7/2}$; evaluating gives $28\pi$. Answer: (C).

   Concept

   An integral of the form $\int_{-2}^{2}x^{4}(4-x^2)^{m}\,dx$ with $m$ a half-integer becomes a Beta/Gamma expression under $x=2\sin\theta$, producing a rational multiple of $\pi$. The choices here ($4\pi,\ \pi/16,\ 28\pi,\ 3\pi/128$) are all multiples of $\pi$, which is only possible when the exponent is $\tfrac{7}{2}$ (a half-integer), so the integrand is read as $x^{4}(4-x^2)^{7/2}$.

   Note

   A strict polynomial power $(4-x^2)^{7}$ would integrate to a rational number with no $\pi$, inconsistent with every option. The exponent is therefore taken as $\tfrac{7}{2}$, which the printed options require; the solution below uses that reading and lands on the exam key.

   Solution

   1. Use symmetry. The integrand is even, so

   $I=\int_{-2}^{2}x^{4}(4-x^2)^{7/2}\,dx=2\int_{0}^{2}x^{4}(4-x^2)^{7/2}\,dx.$

   2. Substitute $x=2\sin\theta$, $dx=2\cos\theta\,d\theta$, with $4-x^2=4\cos^2\theta$:

   $x^{4}=16\sin^4\theta,\qquad (4-x^2)^{7/2}=(4\cos^2\theta)^{7/2}=128\cos^7\theta.$

   Therefore

   $I=2\int_{0}^{\pi/2}16\sin^4\theta\cdot128\cos^7\theta\cdot2\cos\theta\,d\theta=8192\int_{0}^{\pi/2}\sin^4\theta\cos^8\theta\,d\theta.$

   3. Beta function. Using $\int_{0}^{\pi/2}\sin^{2p-1}\theta\cos^{2q-1}\theta\,d\theta=\tfrac12 B(p,q)$ with $p=\tfrac52,\ q=\tfrac92$: …

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10. 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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11. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.Let $m, n, p, q$ be four positive integers. If $\int_{0}^{2\pi} \sin^m x \cos^n x dx = 4 \int_{0}^{\frac{\pi}{2}} \sin^m x \cos^n x dx$, $\int_{0}^{2\pi} \sin^p x \cos^q x dx = 0$, $\int_{0}^{\pi} \sin^r x \cos^q x dx = 0$, $a = m + n + p$ and $b = m + n + q$, then (A) a is even number and b is odd number (B) a is odd number and b is even number (C) Both a and b are even numbers (D) Both a and b are odd numbers

    ›Reveal solutionSolution

    Both $a$ and $b$ are odd numbers — option (D).

    Analyse each condition by symmetry.

    • $\displaystyle\int_{0}^{2\pi}\sin^{m}x\cos^{n}x\,dx=4\int_{0}^{\pi/2}\sin^{m}x\cos^{n}x\,dx$ requires the integrand to be non-negative with quarter-period symmetry, i.e. $m$ and $n$ are both even.
    • $\displaystyle\int_{0}^{2\pi}\sin^{p}x\cos^{q}x\,dx=0$: under $x\to 2\pi-x$ the integral picks up a factor $(-1)^{p}$, so it vanishes only if $p$ is odd. …

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12. 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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