Q.Evaluate the definite integral: $\int_4^5 e^x \ dx$

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.$\displaystyle\int_0^1 (\sqrt{10})^{2x}dx =$ (A) $\dfrac{10}{\log 10}$ (B) $\dfrac{9}{\log 10}$ (C) $\dfrac{1}{\log 10}$ (D) $\dfrac{4}{\log 5}$

   ›Reveal solutionSolution

   Simplify the exponential base first ($(\sqrt{10})^{2x}=10^x$), then integrate directly using the standard exponential-integral formula.

   Concept and Intuition

   $\displaystyle\int a^x\,dx = \dfrac{a^x}{\ln a}+C$ for $a>0,a\neq1$. Recognizing $(\sqrt{10})^{2x}=10^x$ first avoids working with an awkward fractional-power base.

   Step-by-Step Solution

   1. $(\sqrt{10})^{2x} = \left(10^{1/2}\right)^{2x} = 10^x$.
   2. $\displaystyle\int_0^1 10^x\,dx = \left[\dfrac{10^x}{\ln10}\right]_0^1 = \dfrac{10^1-10^0}{\ln10} = \dfrac{10-1}{\ln10} = \dfrac{9}{\ln10}$.

   Common Mistakes …

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2. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/4} e^{\tan^2\theta} \sin^2\theta \tan\theta\, d\theta =$ (A) $\dfrac{1}{2}\left(\dfrac{e}{2} - 1\right)$ (B) $\dfrac{e}{2} - 1$ (C) $\dfrac{\pi}{2}$ (D) $2\left(\dfrac{\pi}{2} - e\right)$

   ›Reveal solutionSolution

   Substituting $v=\tan^2\theta$ converts the integral into $\frac12\int_0^1 \frac{ve^v}{(1+v)^2}\,dv$, which is exactly a product-rule derivative of $\frac{e^v}{1+v}$, giving the clean closed form $\frac12\left(\frac{e}{2}-1\right)$.

   Concept and Intuition

   When an integral mixes $\tan\theta$, $\sec^2\theta$-type factors, and an exponential of $\tan^2\theta$, substituting $v=\tan^2\theta$ (rather than $u=\tan\theta$) often collapses the trigonometric parts neatly, because $dv$ naturally pulls out a $\tan\theta\sec^2\theta$ factor that matches what's needed.

   Step-by-Step Solution

   1. Let $v=\tan^2\theta$. Then $dv = 2\tan\theta\sec^2\theta\,d\theta = 2\tan\theta(1+\tan^2\theta)\,d\theta = 2\tan\theta(1+v)\,d\theta$.
   2. So $\tan\theta\,d\theta = \dfrac{dv}{2(1+v)}$.
   3. Also $\sin^2\theta = \tan^2\theta\cos^2\theta = \dfrac{\tan^2\theta}{1+\tan^2\theta} = \dfrac{v}{1+v}$.
   4. The integrand $e^{\tan^2\theta}\sin^2\theta\tan\theta\,d\theta$ becomes $e^v\cdot\dfrac{v}{1+v}\cdot\dfrac{dv}{2(1+v)} = \dfrac{v\,e^v}{2(1+v)^2}\,dv$.
   5. Limits: as $\theta:0\to\pi/4$, $v=\tan^2\theta: 0\to1$.
   6. So the integral is $\dfrac12\displaystyle\int_0^1 \dfrac{v e^v}{(1+v)^2}\,dv$.
   7. Write $v = (1+v)-1$: $\dfrac{v}{(1+v)^2} = \dfrac1{1+v}-\dfrac1{(1+v)^2}$. …

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3. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\displaystyle\int_{\alpha-1}^{\alpha} \dfrac{e^x(\alpha - x)}{(x-\alpha+1)^2}\,dx =$ (A) $2e^{\alpha} + e$ (B) $\dfrac{2e^{\alpha+2}}{e-2}$ (C) $e^{\alpha}\dfrac{(e+2)}{2}$ (D) $e^{\alpha}\left(\dfrac{e-2}{2}\right)$

   ›Reveal solutionSolution

   This is a shifted $e^x[f(x)+f'(x)]$ integral; substituting $t=x-\alpha$ turns it into the standard $\int \dfrac{t\,e^t}{(t+1)^2}dt = \dfrac{e^t}{t+1}+C$ pattern, evaluated over $t\in[0,1]$.

   Concept and Intuition

   Many definite integrals involving $e^x$ divided by a squared linear factor are disguised applications of the identity $\int e^x\big[g(x)+g'(x)\big]dx = e^x g(x)+C$, which follows straight from the product rule. Recognizing the pattern (a term like $\dfrac{1}{t+1}$ together with its derivative $-\dfrac{1}{(t+1)^2}$, glued together by $e^t$) avoids messy direct integration.

   Step-by-Step Solution

   1. Shift variable: let $t=x-\alpha$, so $x=\alpha+t$, and the range $x:\alpha\to\alpha+1$ becomes $t:0\to1$; the factor $x-\alpha+1$ becomes $t+1$.
   2. The integral becomes $e^\alpha\displaystyle\int_0^1 \dfrac{t\,e^t}{(t+1)^2}\,dt$.
   3. Write $t=(t+1)-1$, so $\dfrac{t}{(t+1)^2}=\dfrac{1}{t+1}-\dfrac{1}{(t+1)^2}$.
   4. Note $\dfrac{d}{dt}\!\left[\dfrac{e^t}{t+1}\right]=\dfrac{e^t(t+1)-e^t}{(t+1)^2}=\dfrac{e^t\,t}{(t+1)^2}$ — exactly the integrand. …

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4. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $\displaystyle\int 4^{x}\,4^{4^{x}}\,4^{4^{4^{x}}}\,dx = A\,4^{4^{4^{x}}}+c$, then $A=$ (A) $\dfrac{1}{\ln 4}$ (B) $\dfrac{1}{(\ln 4)^2}$ (C) $\dfrac{1}{(\ln 4)^3}$ (D) $\dfrac{1}{(\ln 4)^4}$

   ›Reveal solutionSolution

   Repeated application of $\frac{d}{dx}a^{g(x)}=a^{g(x)}\log a\cdot g'(x)$ through a triple exponential tower produces a factor of $(\ln4)^3$, so integrating backwards needs $A=1/(\ln4)^3$.

   Concept and Intuition

   For a tower of exponentials $4^{4^{4^x}}$, differentiating peels one exponential at a time, each time multiplying by $\ln4$ and by the derivative of the next inner exponential. Because the given integrand is exactly the product $4^x\cdot4^{4^x}\cdot4^{4^{4^x}}$ — the "chain" of factors produced by differentiating the tower — this is a reverse-chain-rule integral: recognizing the derivative of the tower directly gives the antiderivative.

   Step-by-Step Solution

   1. Let $F(x)=4^{4^{4^x}}$. Differentiate using the chain rule from the outside in:

   $F'(x)=4^{4^{4^x}}\ln4\cdot\frac{d}{dx}\left(4^{4^x}\right).$

   2. Now $\dfrac{d}{dx}\left(4^{4^x}\right)=4^{4^x}\ln4\cdot\dfrac{d}{dx}(4^x)=4^{4^x}\ln4\cdot 4^x\ln4 = 4^x\cdot4^{4^x}\cdot(\ln4)^2$.
   3. Substituting back: $F'(x)= (\ln4)^3\cdot 4^x\cdot4^{4^x}\cdot4^{4^{4^x}}$. …

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5. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{e^{2x}}{\sqrt[4]{e^x+1}} \, dx =$ (A) $\dfrac{4}{7}(e^x+1)^{4/3}(3e^x-1)+c$ (B) $\dfrac{2}{21}(e^x+1)^{3/4}(3e^x-7)+c$ (C) $\dfrac{4}{21}(e^x+1)^{3/4}(3e^x-4)+c$ (D) $\dfrac{8}{21}(e^x+1)^{3/4}(3e^x-1)+c$

   ›Reveal solutionSolution

   Substitute $u=e^x+1$ to turn this into a simple power-rule integral; simplifies to $\dfrac4{21}(e^x+1)^{3/4}(3e^x-4)+c$ — (C).

   Concept and Intuition

   Writing $e^{2x}=e^x\cdot e^x$ and using $u=e^x+1$ (so $e^x=u-1$, $du=e^xdx$) converts the whole integral into a polynomial-in-$u$ times a fractional power of $u$ — directly integrable term by term.

   Step-by-Step Solution

   1. Let $u=e^x+1\Rightarrow du=e^x\,dx,\ e^x=u-1$.
   2. $e^{2x}dx = e^x\cdot(e^x dx) = (u-1)\,du$.
   3. Integral: $\displaystyle\int\dfrac{(u-1)\,du}{u^{1/4}}=\int\left(u^{3/4}-u^{-1/4}\right)du = \dfrac{u^{7/4}}{7/4}-\dfrac{u^{3/4}}{3/4}+c=\dfrac47u^{7/4}-\dfrac43u^{3/4}+c$.
   4. Factor out $u^{3/4}$: $u^{3/4}\left(\dfrac47u-\dfrac43\right)+c$. Common denominator $21$: $\dfrac47u=\dfrac{12u}{21}$, $\dfrac43=\dfrac{28}{21}$, so this is $u^{3/4}\cdot\dfrac{12u-28}{21}=\dfrac{4}{21}u^{3/4}(3u-7)+c$. …

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6. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.$\displaystyle\int\left(\sum_{r=0}^{\infty} \dfrac{x^r 2^r}{r!}\right)dx =$ (A) $e^x + c$ (B) $\dfrac{-2}{1-2x}+c$ (C) $2e^{2x}+c$ (D) $\dfrac{e^{2x}}{2}+c$

   ›Reveal solutionSolution

   Recognising the Maclaurin series of $e^{2x}$ inside the sum turns this into a one-line integral, giving $\dfrac{e^{2x}}{2}+c$.

   Concept and Intuition

   The exponential series $e^{y}=\sum_{r=0}^\infty \dfrac{y^r}{r!}$ is one of the most important series to recognise instantly. Here $y=2x$, so the given sum is exactly $e^{2x}$ in disguise — the whole problem reduces to a basic integral once you see this.

   Step-by-Step Solution

   1. $\displaystyle\sum_{r=0}^\infty \dfrac{x^r 2^r}{r!} = \sum_{r=0}^\infty \dfrac{(2x)^r}{r!} = e^{2x}$.
   2. So the integral becomes $\displaystyle\int e^{2x}\,dx$.
   3. $\displaystyle\int e^{2x}\,dx = \dfrac{e^{2x}}{2}+c$. …

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7. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.$\int\left(\frac{8^{1+x}+4^{1+x}}{2^{2x}}\right)dx=$ (A) $\frac{2^x}{\log 2}+4x+C$ (B) $8\cdot\frac{2^x}{\log 2}-4x+C$ (C) $8\cdot\frac{2^x}{\log 2}+4x+C$ (D) $\frac{2^x}{\log 2}-4x+C$

   ›Reveal solutionSolution

   Rewrite everything as powers of 2 and simplify the fraction before integrating; the answer is $8\cdot\frac{2^x}{\log 2}+4x+C$.

   Concept and Intuition

   Exponential integrands often simplify dramatically once every term is expressed with the same base. Here base 2 works for 8, 4 and $2^{2x}$ alike, turning a scary-looking fraction into a sum of a simple exponential and a constant.

   Step-by-Step Solution

   1. $8^{1+x}=8\cdot 8^x=8\cdot(2^3)^x=8\cdot 2^{3x}$.
   2. $4^{1+x}=4\cdot 4^x=4\cdot(2^2)^x=4\cdot2^{2x}$.
   3. Divide both by $2^{2x}$: $\dfrac{8\cdot2^{3x}}{2^{2x}}+\dfrac{4\cdot2^{2x}}{2^{2x}}=8\cdot2^{x}+4$.
   4. Integrate term by term: $\int 8\cdot2^x\,dx=8\cdot\dfrac{2^x}{\log2}$, and $\int 4\,dx=4x$. …

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8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$\displaystyle\int \left(\sum_{r=0}^{\infty} \dfrac{x^r 3^r}{r!}\right) dx =$ (A) $e^x + c$ (B) $\dfrac{e^{3x}}{3} + c$ (C) $3e^{3x} + c$ (D) $3e^x + c$

   ›Reveal solutionSolution

   $\sum_{r=0}^\infty \dfrac{(3x)^r}{r!}=e^{3x}$, so the integral is $\dfrac{e^{3x}}{3}+c$.

   Concept and Intuition

   Recognising the exponential series $\sum x^r/r! = e^x$ turns an infinite-series integral into a one-line exponential integration.

   Step-by-Step Solution

   1. $\sum_{r=0}^\infty \dfrac{x^r3^r}{r!} = \sum_{r=0}^\infty \dfrac{(3x)^r}{r!} = e^{3x}$.
   2. $\int e^{3x}\,dx = \dfrac{e^{3x}}{3}+c$.

   Common Mistakes …

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9. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.Assertion (A): $\displaystyle\int_{2}^{e} \left(\dfrac{1}{\log_e x} - \dfrac{1}{(\log_e x)^2}\right) dx = e - 2\log_2 e$ Reason (R): $\displaystyle\int_{a}^{b} e^x\left(f(x) + f'(x)\right) dx = e^b f(b) - e^a f(a)$ (A) (A) and (R) are true, (R) is the correct explanation to (A). (B) (A) and (R) are false, (R) is not the correct explanation to (A). (C) (A) is true and (R) is false, R is not the correct explanation to (A). (D) (A) is false and (R) is true, R is not the correct explanation to (A).

   ›Reveal solutionSolution

   The assertion is a direct application of the reason's identity $\int e^x(f+f')dx=e^xf(x)+C$ under the substitution $x=e^t$; both statements check out and R explains A.

   Concept and Intuition

   The identity $\displaystyle\int_a^b e^x\big(f(x)+f'(x)\big)dx = e^bf(b)-e^af(a)$ comes from noticing $\frac{d}{dx}\left[e^xf(x)\right]=e^x(f(x)+f'(x))$ — a product-rule pattern. Many integrals disguise this pattern after a substitution.

   Step-by-Step Solution

   1. In the assertion's integral, substitute $x=e^t \Rightarrow dx = e^t\,dt$ and $\ln x = t$. Limits: $x=2\Rightarrow t=\ln2$; $x=e\Rightarrow t=1$.
   2. The integral becomes $\displaystyle\int_{\ln2}^{1}\left(\frac1t-\frac1{t^2}\right)e^t\,dt$.
   3. Let $f(t)=\dfrac1t$, so $f'(t)=-\dfrac1{t^2}$. The integrand is exactly $e^t\big(f(t)+f'(t)\big)$, matching the Reason's identity with $a=\ln2,\ b=1$.
   4. Apply the Reason: value $= e^{1}f(1) - e^{\ln2}f(\ln2) = e\cdot 1 - 2\cdot\dfrac{1}{\ln2}$. …

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10. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
    Q.If $\displaystyle\int \dfrac{a^{x+1}}{a^x + a^{-x}}\, dx = A\log(a^x + a^{-x}) + Bx + C$ then $\dfrac{A}{B} =$ (A) $\log_a e$ (B) $\log_e a$ (C) $\dfrac{a}{\log_e a}$ (D) $\dfrac{a}{\log_a e}$

    ›Reveal solutionSolution

    Simplifying via $u=a^{2x}+1$ and re-expressing in terms of $a^x+a^{-x}$ gives $A/B = \log_a e$.

    Concept and Intuition

    The trick with $\int \dfrac{a^{x+1}}{a^x+a^{-x}}dx$ is to multiply through by $a^x/a^x$ so the denominator becomes a polynomial in $a^{2x}$, turning it into a standard $\int \dfrac{du}{u}$ form after substitution. The subtlety is that this substitution naturally produces $\ln(a^{2x}+1)$, which must be related back to the given target form $\ln(a^x+a^{-x})$ using the identity $a^{2x}+1 = a^x(a^x+a^{-x})$.

    Step-by-Step Solution

    1. Multiply numerator and denominator by $a^x$: $\dfrac{a\cdot a^x}{a^x+a^{-x}} = \dfrac{a\cdot a^{2x}}{a^{2x}+1}$.
    2. Let $u = a^{2x}+1$, so $du = 2a^{2x}\ln a\,dx$, i.e. $a^{2x}dx = \dfrac{du}{2\ln a}$.
    3. Integral becomes $a\int \dfrac{1}{u}\cdot\dfrac{du}{2\ln a} = \dfrac{a}{2\ln a}\ln|u| + C = \dfrac{a}{2\ln a}\ln(a^{2x}+1)+C$.
    4. Use $a^{2x}+1 = a^x(a^x+a^{-x})$, so $\ln(a^{2x}+1) = \ln(a^x+a^{-x}) + x\ln a$. …

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11. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.If $\displaystyle\int \dfrac{e^{\sqrt{x}}}{\sqrt{x}}\left(x + \sqrt{x}\right)dx = e^{\sqrt{x}}\left[Ax + B\sqrt{x} + C\right] + K$, then $A + B + C =$ (A) -2 (B) 2 (C) 4 (D) -4

    ›Reveal solutionSolution

    Substituting $t=\sqrt x$ converts the integral into a standard $\int e^t(t^2+t)\,dt$, which evaluates to $e^{\sqrt x}(2x-2\sqrt x+2)+K$, so $A+B+C=2-2+2=2$.

    Concept and Intuition

    Integrals with $e^{\sqrt x}$ and $\sqrt x$ everywhere are cleaned up by the substitution $t=\sqrt x$, turning them into polynomial-times-exponential integrals that use the standard reduction formula $\int t^ne^t\,dt$.

    Step-by-Step Solution

    1. Let $t=\sqrt x \Rightarrow x=t^2,\ dx=2t\,dt$.
    2. Integral: $\displaystyle\int\dfrac{e^t}{t}(t^2+t)\cdot2t\,dt = \int e^t\cdot t(t+1)\cdot2\,dt = 2\int e^t(t^2+t)\,dt$.
    3. Use $\int t^2e^t\,dt = e^t(t^2-2t+2)$ and $\int te^t\,dt=e^t(t-1)$ (both by repeated integration by parts).
    4. Sum: $e^t(t^2-2t+2)+e^t(t-1) = e^t(t^2-t+1)$.
    5. Multiply by $2$: $2e^t(t^2-t+1) = e^t(2t^2-2t+2)$. …

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12. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{3e^{x} - 7e^{-x}}{7e^{x} + 3e^{-x}}\,dx = Kx + L\log\left(e^{-2x} + \dfrac{7}{3}\right) + C$, then $K + L =$ (A) $\dfrac{-3}{38}$ (B) $\dfrac{21}{38}$ (C) $\dfrac{38}{21}$ (D) $\dfrac{-38}{3}$

    ›Reveal solutionSolution

    Splitting the numerator as a combination of the denominator and its derivative gives $Kx+L\log D$; converting $D$ to the $e^{-2x}+7/3$ form shows $K+L=\dfrac{38}{21}$.

    Concept and Intuition

    For an integral of the form $\int\dfrac{N(x)}{D(x)}dx$ where $N,D$ are combinations of $e^x,e^{-x}$, the standard trick is to write $N = A\,D + B\,D'$. Then $\int\dfrac{N}{D}dx = A\int dx + B\int\dfrac{D'}{D}dx = Ax+B\log|D|+C$ — turning a messy rational-exponential integral into an elementary one.

    Step-by-Step Solution

    1. Let $D = 7e^x+3e^{-x}$, so $D' = 7e^x-3e^{-x}$.
    2. Seek $A,B$ with $3e^x-7e^{-x} = A(7e^x+3e^{-x})+B(7e^x-3e^{-x})$.
    3. Coefficient of $e^x$: $7A+7B=3$. Coefficient of $e^{-x}$: $3A-3B=-7$.
    4. Solve: $A+B=\dfrac37$, $A-B=-\dfrac73$. Adding: $2A = \dfrac37-\dfrac73=\dfrac{9-49}{21}=-\dfrac{40}{21}\Rightarrow A=-\dfrac{20}{21}$. Subtracting: $2B=\dfrac37+\dfrac73=\dfrac{9+49}{21}=\dfrac{58}{21}\Rightarrow B=\dfrac{29}{21}$.
    5. So $\int\frac{N}{D}dx = Ax + B\log D + C$, i.e. using $D=7e^x+3e^{-x}=e^x\left(7+3e^{-2x}\right)=3e^x\left(e^{-2x}+\tfrac73\right)$: $\log D = x+\log 3+\log\left(e^{-2x}+\tfrac73\right)$. …

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