Q.Evaluate the definite integral: $\int_{-1}^1 (x+1) \ dx$

Showing the 12 most recent of 22 on this concept.

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

   4. Add the results

   $-\frac{1}{3} + \frac{1}{3} = 0.$

   Tip

   Recognizing odd/even symmetry saves time. For any odd function $f$ (i.e., $f(-x) = -f(x)$), $\int_{-a}^{a} f(x) \, dx = 0$ automatically. This is a classic shortcut for symmetric integrals.

   Watch out

   A common mistake is to forget the sign change for negative $x$ when dealing with absolute values. Always check the piecewise definition before integrating.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

2. 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

   $2I = \int_0^1 1\,dx = \bigl[x\bigr]_{0}^{1} = 1.$

   4. Solve for $I$.

   $I = \frac12.$

   Watch out

   A common mistake is to assume $f$ is constant. The condition only forces the sum at symmetric points to be $1$, not that $f$ itself is constant. For example, $f(x)=x$ works because $x+(1-x)=1$, and its integral is indeed $\frac12$. The method above works for any function satisfying the condition.

   Tip

   This is a classic “symmetric sum” trick: whenever you see $f(x)+f(a-x)=c$, the integral over $[0,a]$ is $\frac{c a}{2}$. Here $a=1$, $c=1$, so the answer is $\frac12$ immediately.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. 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

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

   Watch out

   A common mistake is to forget the sign from $dx = -dt$ or to mishandle the limits. Always check: substituting $x = -t$ gives $dx = -dt$, and the limits reverse; the two negatives (one from $dt$, one from swapping limits) cancel cleanly.

   Tip

   This trick — splitting an integral over a symmetric interval and substituting $x \to -x$ on the negative part — is a standard technique for integrals of odd/even functions or functions with logarithmic singularities. Here it reveals that the original integral from $-1$ to $1$ equals $\int_0^1 \frac{\log(1+x) + \log(1-x)}{1+x^2} dx$, which simplifies further using $\log(1-x^2)$.

   ✓Final answer

   The correct option is (C): $f(x) = \frac{\log(1-x)}{1+x^2}$.

4. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.$\lim_{n \to \infty} \frac{(2n(2n-1)\ldots(n+2)(n+1))^{1/n}}{n} =$ (A) $\int_{0}^{1} \log x \, dx$ (B) $\int_{0}^{1} (x+1)\log(x+1) \, dx$ (C) $\int_{0}^{1} \log(1+x) \, dx$ (D) $\int_{0}^{1} x \log x \, dx$

   ›Reveal solutionSolution

   The product is $\prod_{k=1}^{n}(n+k)$; taking $\tfrac1n$-th power and dividing by $n$ turns the log into a Riemann sum for $\displaystyle\int_0^1\log(1+x)\,dx$. Answer (C).

   Rewrite the product

   $2n(2n-1)\cdots(n+2)(n+1)=\prod_{k=1}^{n}(n+k).$

   So the limit is

   $L=\lim_{n\to\infty}\frac{1}{n}\left(\prod_{k=1}^{n}(n+k)\right)^{1/n} =\lim_{n\to\infty}\left(\prod_{k=1}^{n}\frac{n+k}{n}\right)^{1/n} =\lim_{n\to\infty}\left(\prod_{k=1}^{n}\Bigl(1+\frac{k}{n}\Bigr)\right)^{1/n}.$

   Take logarithms

   $\log L=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\log\!\Bigl(1+\frac{k}{n}\Bigr)=\int_{0}^{1}\log(1+x)\,dx.$

   This is exactly option (C). (Evaluating, $\displaystyle\int_0^1\log(1+x)\,dx=2\log 2-1$, so $L=\dfrac{4}{e}$.)

   ✓Final answer

   $\log L=\displaystyle\int_{0}^{1}\log(1+x)\,dx$ — option (C).

5. 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:

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

   The denominators are identical, so the sum of numerators is $\sqrt{x} + \sqrt{9-x}$, which cancels the denominator:

   $2I = \int_{3}^{6} 1 \, dx.$

   4. Evaluate the simple integral

   $\int_{3}^{6} 1 \, dx = 6 - 3 = 3.$

   Hence $2I = 3$, so $I = \frac{3}{2}$.

   Tip

   This trick works whenever the integrand is of the form $\frac{f(x)}{f(x)+f(a+b-x)}$ over $[a,b]$. The result is always $\frac{b-a}{2}$, independent of $f$ (as long as $f$ is positive and integrable).

   Watch out

   A common mistake is to try direct substitution or partial fractions — that leads to messy algebra. The symmetry method is far cleaner and avoids any heavy computation.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

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

   Combining over a common denominator $35$:

   $\frac{1792}{35} - \frac{5600}{35} + \frac{3840}{35} - \frac{1120}{35} = \frac{-1088}{35}+\frac{4832}{35}=\frac{32}{35}.$

   ✓Final answer

   $\displaystyle\int_{0}^{2} x^3(2-x)^4\,dx = \dfrac{32}{35}$ — option (D).

7. 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 $\pi/2$:

   $\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^3 x \, dx = 2 \int_{0}^{\pi/2} \sin^2 x \cos^3 x \, dx.$

   Use substitution $u = \sin x$, $du = \cos x \, dx$. Then $\cos^2 x = 1 - u^2$, so $\cos^3 x = (1-u^2)\cos x$. The integral becomes:

   $2 \int_{0}^{1} u^2 (1-u^2) \, du = 2 \int_{0}^{1} (u^2 - u^4) \, du = 2\left[ \frac{u^3}{3} - \frac{u^5}{5} \right]_0^1 = 2\left( \frac{1}{3} - \frac{1}{5} \right) = 2 \cdot \frac{2}{15} = \frac{4}{15}.$

   5. Combine results: Total integral = $0 + \frac{4}{15} = \frac{4}{15}$.

   Watch out

   A common mistake is to think the whole integrand is odd because $\sin x + \cos x$ looks “odd-ish,” but $\cos x$ is even. Always check each term separately.

   Tip

   Even functions can be integrated from $0$ to $a$ and doubled; odd functions vanish. This saves half the work.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

8. 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 $0\to\pi/2$:

   $I=\frac14\int_0^{\pi/2}\sin^4u\,du=\frac14\cdot\frac{3\pi}{16}=\frac{3\pi}{64},$

   using the Wallis value $\displaystyle\int_0^{\pi/2}\sin^4u\,du=\frac{3\cdot1}{4\cdot2}\cdot\frac{\pi}{2}=\frac{3\pi}{16}$.

   ✓Final answer

   $\displaystyle\int_{-\pi/8}^{\pi/8}\frac{\sin^4(4x)}{1+e^{4x}}\,dx=\frac{3\pi}{64}$ — option (C).

9. 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.

   Watch out

   A common mistake is to think the answer is $2\int_{0}^{a} f(x) dx$ (option A). That would only be true if $f(x) = f(x+a)$ for all $x$ in $[0,a]$, i.e., if $f$ is periodic with period $a$. The problem gives no such condition, so (A) is not generally correct.

   Tip

   This property is extremely useful: whenever you see an integral from $0$ to $2a$, try splitting and shifting. It often simplifies problems where $f(x)$ and $f(x+a)$ have a known relationship, like $f(x) + f(x+a) = \text{constant}$.

   Now check the other options briefly:

   • Option (C) says the integral is $0$, which is false for a general function (e.g., $f(x)=1$ gives $2a \neq 0$).
   • Option (D) says $\int_{0}^{2a} f(2a + x) dx$. Let $u = 2a + x$, then $dx = du$, limits $x=0 \to u=2a$, $x=2a \to u=4a$, so this equals $\int_{2a}^{4a} f(u) du$, which is not generally equal to the original integral.

   ✓Final answer

   The correct option is (B): $\int_{0}^{2a} f(x) dx = \int_{0}^{a} (f(x) + f(x+a)) dx$.

10. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.If $f$ is defined on $\mathbb{R}$ such that $f(x)f(-x)=9$, then $\int_{-23}^{23} \frac{dx}{3+f(x)}=$ (A) $\frac{51}{3}$ (B) $\frac{49}{3}$ (C) $\frac{46}{3}$ (D) $\frac{46}{6}$

    ›Reveal solutionSolution

    Using the symmetry $f(x)f(-x)=9$ together with the substitution $x\to -x$, the integrand collapses to a constant. The paired integral gives $2I=\tfrac{46}{3}$, so $I=\tfrac{46}{6}$.

    Setting up the symmetry. Let

    $I=\int_{-23}^{23}\frac{dx}{3+f(x)}.$

    Applying the substitution $x\to -x$ (which leaves the limits $-23$ to $23$ unchanged) gives an equal value:

    $I=\int_{-23}^{23}\frac{dx}{3+f(-x)}.$

    Using the given relation. Since $f(x)f(-x)=9$, we have $f(-x)=\dfrac{9}{f(x)}$. Substituting,

    $I=\int_{-23}^{23}\frac{dx}{3+\dfrac{9}{f(x)}}=\int_{-23}^{23}\frac{f(x)\,dx}{3f(x)+9}=\int_{-23}^{23}\frac{f(x)}{3\,[\,3+f(x)\,]}\,dx.$

    Adding the two forms of $I$.

    $2I=\int_{-23}^{23}\left[\frac{1}{3+f(x)}+\frac{f(x)}{3\,[\,3+f(x)\,]}\right]dx=\int_{-23}^{23}\frac{3+f(x)}{3\,[\,3+f(x)\,]}\,dx=\int_{-23}^{23}\frac{1}{3}\,dx.$

    The integrand is now the constant $\tfrac13$ over an interval of length $46$:

    $2I=\frac{1}{3}\times 46=\frac{46}{3}\quad\Rightarrow\quad I=\frac{46}{6}.$

    ✓Final answer

    $\displaystyle\int_{-23}^{23}\frac{dx}{3+f(x)}=\frac{46}{6}$ — option (D).

11. 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.

    $I=\int_{0}^{\pi/15}\cos 5x\,dx=\left[\frac{\sin 5x}{5}\right]_0^{\pi/15}=\frac{1}{5}\sin\!\left(\frac{\pi}{3}\right)=\frac{1}{5}\cdot\frac{\sqrt3}{2}=\frac{\sqrt3}{10}.$

    ✓Final answer

    $\displaystyle\int_{-\pi/15}^{\pi/15}\frac{\cos 5x}{1+e^{5x}}\,dx=\frac{\sqrt3}{10}$ — option (B).

12. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
    Q.Let $f:\mathbb{R} \to \mathbb{R}$ be an odd function and $\int_{-1}^{1} x^3 f''(x) dx = \frac{8}{5}$. If $g(x) = x f(x)$, $g'(1) = \frac{8}{3}$ and $\int_{0}^{1} g(x) dx = \frac{2}{15}$, then $f(1) =$ (A) $\frac{1}{2}$ (B) $\frac{2}{3}$ (C) $\frac{3}{4}$ (D) $\frac{4}{5}$

    ›Reveal solutionSolution

    Using integration by parts and the given oddness of $f$, we reduce the integral condition to an equation involving $f(1)$ and $g'(1)$, then solve to find $f(1) = \frac{2}{3}$.

    We are told $f$ is odd: $f(-x) = -f(x)$. This implies $f(0)=0$ and $f'$ is even, $f''$ is odd. The function $g(x) = x f(x)$ is even (product of odd and odd), so $g$ is even. That symmetry will simplify integrals.

    We have three pieces of data:

    1. $\int_{-1}^{1} x^3 f''(x) \, dx = \frac{8}{5}$
    2. $g'(1) = \frac{8}{3}$
    3. $\int_{0}^{1} g(x) \, dx = \frac{2}{15}$

    We want $f(1)$.

    ---

    1. Use integration by parts on the first integral. Since $f''$ is odd and $x^3$ is odd, their product is even, so

    $\int_{-1}^{1} x^3 f''(x) \, dx = 2 \int_{0}^{1} x^3 f''(x) \, dx = \frac{8}{5}.$

    Hence

    $\int_{0}^{1} x^3 f''(x) \, dx = \frac{4}{5}.$

    Now integrate by parts: let $u = x^3$, $dv = f''(x) dx$, so $du = 3x^2 dx$, $v = f'(x)$. Then

    $\int_{0}^{1} x^3 f''(x) dx = \left[ x^3 f'(x) \right]_{0}^{1} - \int_{0}^{1} 3x^2 f'(x) dx.$

    At $x=0$, the term vanishes; at $x=1$, we get $1^3 f'(1) = f'(1)$. So

    $\frac{4}{5} = f'(1) - 3 \int_{0}^{1} x^2 f'(x) dx. \qquad(1)$

    2. Relate $g'(1)$ to $f$ and $f'$. Since $g(x) = x f(x)$, differentiate:

    $g'(x) = f(x) + x f'(x).$

    At $x=1$,

    $g'(1) = f(1) + f'(1) = \frac{8}{3}. \qquad(2)$

    3. Use the integral of $g$.

    $\int_{0}^{1} g(x) dx = \int_{0}^{1} x f(x) dx = \frac{2}{15}. \qquad(3)$

    4. Connect the integral in (1) to the known integral (3). Integrate $\int x^2 f'(x) dx$ by parts: let $u = x^2$, $dv = f'(x) dx$, so $du = 2x dx$, $v = f(x)$. Then

    $\int_{0}^{1} x^2 f'(x) dx = \left[ x^2 f(x) \right]_{0}^{1} - \int_{0}^{1} 2x f(x) dx.$

    At $x=0$, term is $0$; at $x=1$, we get $1^2 f(1) = f(1)$. So

    $\int_{0}^{1} x^2 f'(x) dx = f(1) - 2 \int_{0}^{1} x f(x) dx.$

    Using (3), $\int_{0}^{1} x f(x) dx = \frac{2}{15}$, so

    $\int_{0}^{1} x^2 f'(x) dx = f(1) - 2 \cdot \frac{2}{15} = f(1) - \frac{4}{15}. \qquad(4)$

    5. Substitute (4) into (1). From (1):

    $\frac{4}{5} = f'(1) - 3 \left( f(1) - \frac{4}{15} \right).$

    Simplify:

    $\frac{4}{5} = f'(1) - 3 f(1) + \frac{4}{5}.$

    Cancel $\frac{4}{5}$ from both sides, giving

    $0 = f'(1) - 3 f(1) \quad \Rightarrow \quad f'(1) = 3 f(1). \qquad(5)$

    6. Combine with (2) to solve for $f(1)$. From (2): $f(1) + f'(1) = \frac{8}{3}$. Substitute $f'(1) = 3 f(1)$:

    $f(1) + 3 f(1) = \frac{8}{3} \quad \Rightarrow \quad 4 f(1) = \frac{8}{3} \quad \Rightarrow \quad f(1) = \frac{2}{3}.$

    Tip

    The cancellation of $\frac{4}{5}$ is not a coincidence — it’s forced by the consistency of the given data. This is a neat check that the numbers were chosen to work out cleanly.

    Watch out

    A common mistake is forgetting that $f$ is odd, which makes $x^3 f''(x)$ even, so the integral from $-1$ to $1$ is twice the integral from $0$ to $1$. Without this, the factor of 2 is lost and the answer changes.

    ✓Final answer

    The correct option is (B).

    ANSWER: B