Integrals - Test Papers

 CBSE Test Paper 01

Chapter 7 Integrals

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1. ∫1ex+1dx$\int \frac{1}{e^x+1}dx$ is equal to 

   1. log (1+e−2x)+C$(1+e^{-2x})+C$
   2. log(e−2x−2x)+C$\log (e^{-2x}-2x)+C$
   3. – log(1 + e-x) + C
   4. log(e3x+x)+C$\log (e^{3x}+x)+C$

2. The functionf(x)=∫0xlog(t+1+t2)−−−−−−√dt$f(x)=\underset{0}{\overset{x}{\int }}\log (t+\sqrt{1+t^2)}dt$ is 

   1. an odd function
   2. an even function
   3. Neither odd nor Even
   4. a periodic function

3. ∫0π/2log|cosx|$\underset{0}{\overset{\pi /2}{\int }}\log \left|\cos x\right|$dx is equal to 

   1. −π2log2$-\frac{\pi }{2}\log 2$
   2. πlog 2$\pi log2$
   3. π2log3$\frac{\pi }{2}\log 3$
   4. −π3log3$-\frac{\pi }{3}\log 3$

4. ∫ablogxxdx$\underset{a}{\overset{b}{\int }}\frac{\log x}{x}dx$ is equal to 

   1. log(b−a)b−a$\frac{log(b-a)}{b-a}$
   2. log (a+b).log (b–a)
   3. log(ab).log(ba)$\log (ab).log(\frac{b}{a})$
   4. 12log(ab).log(ba)$\frac{1}{2}\log (ab).log\left(\frac{b}{a}\right)$

5. If ∫f(x)$\int f(x)$ dx = f (x), then 

   1. f(x) = ax
   2. f (x) = x
   3. f(x) = 0
   4. f(x)=ex$f(x)=e^x$
6. The function A(x) denotes the ________ function and is given by A (x) = ∫axf(x)dx$\underset{a}{\overset{x}{\int }}f(x)dx$.
7. The indefinite integral of 2x12$2x^{\frac{1}{2}}$ is ________.
8. The indefinite integral of 2x2 + 3 is ________.

9. Show that ∫2x+3x2+3xdx=log∣∣x2+3x∣∣+C$\int \frac{2x+3}{x^2+3x}dx=\log \left|x^2+3x\right|+C$. 

10. Evaluate ∫1011−x2√dx.${\int }_0^1\frac{1}{\sqrt{1-x^2}}dx.$ 

11. Evaluate ∫sin3xdx.$\int {\sin }^{3}xdx.$ 

12. Evaluate the definite integral ∫0π4(2sec2x+x3+2)dx$\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(2{\sec }^{2}x+x^3+2\right)dx$. 

13. Integrate the following function 17−6x−x2√$\frac{1}{\sqrt{7-6x-x^2}}$. 

14. Integrate the function (x2 + 1) log x 

15. ∫π/40sinxcosxcos4x+sin4xdx${\int }_0^{\pi /4}\frac{\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$. 

16. Evaluate ∫1−sinx√1+cosxe−x2dx.$\int \frac{\sqrt{1-\sin x}}{1+\cos x}{e}^{\frac{-x}{2}}dx.$ 

17. Evaluate ∫01xlog(1+2x)dx$\underset{0}{\overset{1}{\int }}x\log \left(1+2x\right)dx$ 

18. Evaluate ∫31(2x2+5x)dx${\int }_1^3\left(2x^2+5x\right)dx$ as a limit of a sum. 

CBSE Test Paper 01
Chapter 7 Integrals

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Solution

1. 3. –log(1 + e-x) + C, Explanation: ∫e−x1+e−xdx=−∫−e−x1+e−xdx=−log(1+e−x)+C$\int \frac{e^{-x}}{1+e^{-x}}dx=-\int \frac{-e^{-x}}{1+e^{-x}}dx=-\log (1+e^{-x})+C$
2. 2. an even function, Explanation: t = -u
      f(-x) = ∫0xlog(−u+1+u2−−−−−√)(−du)$\underset{0}{\overset{x}{\int }}\log (-u+\sqrt{1+u^2})(-du)$
      = −∫0πlog(1+u+u21+u2+u√)(−du)$-\underset{0}{\overset{\pi }{\int }}\log \left(\frac{1+u+u^2}{\sqrt{1+u^2+u}}\right)(-du)$
      = ∫0xlog(u+1+u2−−−−−√)du$\underset{0}{\overset{x}{\int }}\log (u+\sqrt{1+u^2})du$ = f(x)
      ⇒$\Rightarrow$ f(-x) = f(x) ⇒$\Rightarrow$ f is an even function
3. 1. −π2log2$-\frac{\pi }{2}\log 2$, Explanation: ∫0π/2log|cosx|dx$\underset{0}{\overset{\pi /2}{\int }}\log \left|\cos x\right|dx$
      =∫0π/2log∣∣cos(π2−x)∣∣dx$=\underset{0}{\overset{\pi /2}{\int }}\log \left|\cos (\frac{\pi }{2}-x)\right|dx$
      =∫0π/2log|sinx|dx$=\underset{0}{\overset{\pi /2}{\int }}\log \left|\sin x\right|dx$
      =−π2log2$=-\frac{\pi }{2}\log 2$(standard result)
4. 4. 12log(ab).log(ba)$\frac{1}{2}\log (ab).log\left(\frac{b}{a}\right)$, Explanation: =∫ab(logx)1(1x)dx$=\underset{a}{\overset{b}{\int }}{{(logx)}^{{}^1}}^{}(\frac{1}{x})dx$
      (letlogx=tthen1xdx=dt)$(letlogx=tthen\frac{1}{x}dx=dt)$
      =[(logx)22]ba$={\left[\frac{{(logx)}^2}{2}\right]}_a^b$
      =12[(logb)2−(loga)2]$=\frac{1}{2}\left[{(logb)}^2-{(loga)}^2\right]$
      =12(logb+loga)(logb−loga)$=\frac{1}{2}(logb+loga)(logb-loga)$
      =12log(ab)logba$=\frac{1}{2}\log (ab)\log \frac{b}{a}$
5. 4. f(x)=ex$f(x)=e^x$, mExplanation: ddx(f(x))=f(x)$\frac{d}{dx}(f(x))=f(x)$
      It implies that the function remains same after integrating or differentiating it. So the function must be ex
6. area
7. 43x32+c$\frac{4}{3}{x}^{\frac{3}{2}}+c$
8. =45x54+c$=\frac{4}{5}{x}^{\frac{5}{4}}+c$
9. Let I=∫2x+3x2+3xdx$I=\int \frac{2x+3}{x^2+3x}dx$
   Put x2 + 3x = t
   ⇒$\Rightarrow$ (2x + 3)dx = dt
   ∴I=∫1tdt=log|t|+C$\therefore I=\int \frac{1}{t}dt=\log \left|t\right|+C$
   = log |(x2 + 3x)| + C
10. According to the question , I=∫1011−x2√dx$I={\int }_0^1\frac{1}{\sqrt{1-x^2}}dx$
    = [sin−1x]10[∵∫ab11−x2√dx=sin−1(a)−sin−1(b)]${\left[{\sin }^{-1}x\right]}_0^1\left[\because {\int }_b^a\frac{1}{\sqrt{1-x^2}}dx={\sin }^{-1}(a)-{\sin }^{-1}(b)\right]$
    =sin−1(1)−sin−1(0)$=si{n}^{-1}(1)-si{n}^{-1}(0)$
    =sin−1(sinπ2)−sin−1(sin0)$={\sin }^{-1}\left(\sin \frac{\pi }{2}\right)-{\sin }^{-1}(\sin 0)$
    =π2−0=π2$=\frac{\pi }{2}-0=\frac{\pi }{2}$
11. Let I=∫sin3xdx=∫3sinx−sin3x4dx$I=\int {\sin }^{3}xdx=\int \frac{3\sin x-\sin 3x}{4}dx$ [∵sin3x=3sinx−4sin3x]$\left[\because \sin 3x=3\sin x-4{\sin }^{3}x\right]$
    =14∫3sinxdx−14∫sin3xdx$=\frac{1}{4}\int 3\sin xdx-\frac{1}{4}\int \sin 3xdx$
    =14(−3cosx+cos3x3)+C$=\frac{1}{4}\left(-3\cos x+\frac{\cos 3x}{3}\right)+C$
12. ∫0π4(2sec2x+x3+2)dx$\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(2{\sec }^{2}x+x^3+2\right)dx$
    =2∫0π4sec2xdx+∫0π4x3dx+2∫0π41dx$=2\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\sec }^{2}xdx+\underset{0}{\overset{\frac{\pi }{4}}{\int }}{x}^{3}dx+2\underset{0}{\overset{\frac{\pi }{4}}{\int }}1dx$
    =2(tanx)π40+(x44)π40+2(x)π40$=2{\left(\tan x\right)}_0^{\frac{\pi }{4}}+{\left(\frac{x^4}{4}\right)}_0^{\frac{\pi }{4}}+2{\left(x\right)}_0^{\frac{\pi }{4}}$
    =2(tanπ4−tan0o)+(π4)44−0$=2\left(\tan \frac{\pi }{4}-\tan 0^o\right)+\frac{{\left(\frac{\pi }{4}\right)}^4}{4}-0$ +2(π4−0)$+2\left(\frac{\pi }{4}-0\right)$
    =2(1−0)+(π4256)4+2π4$=2\left(1-0\right)+\frac{\left(\frac{{\pi }^4}{256}\right)}{4}+\frac{2\pi }{4}$
    =2+π41024+π2$=2+\frac{{\pi }^4}{1024}+\frac{\pi }{2}$
    =π41024+π2+2$=\frac{{\pi }^4}{1024}+\frac{\pi }{2}+2$
13. ∫17−6x−x2√dx$\int \frac{1}{\sqrt{7-6x-x^2}}dx$
    =∫1−x2−6x+7√dx$=\int \frac{1}{\sqrt{-x^2-6x+7}}dx$
    =∫1−(x2+6x−7)√dx$=\int \frac{1}{\sqrt{-\left(x^2+6x-7\right)}}dx$
    =∫1−(x2+6x+9−9−7)√dx$=\int \frac{1}{\sqrt{-\left(x^2+6x+9-9-7\right)}}dx$
    =∫1−{(x+3)2−16}√dx$=\int \frac{1}{\sqrt{-\left\{{\left(x+3\right)}^2-16\right\}}}dx$
    =∫1(16)−(x+3)2√dx$=\int \frac{1}{\sqrt{\left(16\right)-{\left(x+3\right)}^2}}dx$
    =∫1(4)2−(x+3)2√dx$=\int \frac{1}{\sqrt{{\left(4\right)}^2-{\left(x+3\right)}^2}}dx$
    =sin−1(x+34)+c$={\sin }^{-1}\left(\frac{x+3}{4}\right)+c$
    [∵∫1a2−x2dx=sin−1xa]$\left[\because \int \frac{1}{a^2-x^2}dx={\sin }^{-1}\frac{x}{a}\right]$
14. ∫(x2+1)logxdx$\int \left(x^2+1\right)\log xdx$
    =∫(logx)(x2+1)dx$=\int \left(\log x\right)\left(x^2+1\right)dx$
    [Applying product rule]
    =logx(x33+x)−∫1x(x33+x)dx$=\log x\left(\frac{x^3}{3}+x\right)-\int \frac{1}{x}\left(\frac{x^3}{3}+x\right)dx$
    =(x33+x)logx−∫(x23+1)dx$=\left(\frac{x^3}{3}+x\right)\log x-\int \left(\frac{x^2}{3}+1\right)dx$
    =(x33+x)logx−13∫x2dx−∫1dx$=\left(\frac{x^3}{3}+x\right)\log x-\frac{1}{3}\int x^{2}dx-\int 1dx$
    =(x33+x)logx−13x33−x+c$=\left(\frac{x^3}{3}+x\right)\log x-\frac{1}{3}\frac{x^3}{3}-x+c$
    =(x33+x)logx−x39−x+c$=\left(\frac{x^3}{3}+x\right)\log x-\frac{x^3}{9}-x+c$
15. I=∫π/40sinxcosxcos4x+sin4xdx$I={\int }_0^{\pi /4}\frac{\sin x\cos x}{{\cos }^{4}x+{\sin }^{4}x}dx$
    Dividing Numerator and Denominator by cos4x${\cos }^{4}x$
    =∫π/40sinx.cosxcos4xcos4xcos4x+sin4xcos4xdx$={\int }_0^{\pi /4}\frac{\frac{\sin x.\cos x}{{\cos }^{4}x}}{\frac{{\cos }^{4}x}{{\cos }^{4}x}+\frac{{\sin }^{4}x}{{\cos }^{4}x}}dx$
    =∫π/40tanx.sec2x1+tan4xdx$={\int }_0^{\pi /4}\frac{\tan x.{\sec }^{2}x}{1+{\tan }^{4}x}dx$
    =∫π/40tanx.sec2x1+(tan2x)2dx$={\int }_0^{\pi /4}\frac{\tan x.{\sec }^{2}x}{1+{\left({\tan }^{2}x\right)}^2}dx$
    put tan2x=t${\tan }^{2}x=t$
    2tanx.sec2xdx=dt$2\tan x.{\sec }^{2}xdx=dt$
    when x=0, t=0 and when x=π4$\frac{\pi }{4}$ t=1
    ∴I=12∫10dt1+t2$\therefore I=\frac{1}{2}{\int }_0^1\frac{dt}{1+t^2}$
    =12[tan−1t]10$=\frac{1}{2}{\left[{\tan }^{-1}t\right]}_0^1$
    =12.π4=π8$=\frac{1}{2}.\frac{\pi }{4}=\frac{\pi }{8}$
16. Given, I=∫1−sinx√1+cosx⋅e−x2dx$I=\int \frac{\sqrt{1-\sin x}}{1+\cos x}\cdot e^{\frac{-x}{2}}dx$
    Let −x2=t⇒dx=−2dt$\frac{-x}{2}=t\Rightarrow dx=-2dt$
    I=∫1−sin(−2t)√1+cos(−2t)et(−2dt)$I=\int \frac{\sqrt{1-\sin (-2t)}}{1+\cos (-2t)}{e}^t(-2dt)$ [∵x=−2t]$[\because x=-2t]$
    =−2∫et1+sin2t√1+cos2tdt$=-2\int e^t\frac{\sqrt{1+\sin 2t}}{1+\cos 2t}dt$[∵sin(−θ)=−sinθ and cos(−θ)=cosθ]$[\because \sin (-\theta )=-\sin \theta \text{ and }\cos (-\theta )=\cos \theta ]$
    =−2∫etcos2x+sin2x+2sinxcosx√1+cos2tdt[∵cos2x+sin2x=1,sin2x=2sinxcosx]$=-2\int e^t\frac{\sqrt{co{s}^{2}x+si{n}^{2}x+2sinxcosx}}{1+\cos 2t}dt[\because co{s}^{2}x+si{n}^{2}x=1,sin2x=2sinxcosx]$
    =−2∫et((cost+sint)2√2cos2t)dt[∵(a+b)2=a2+b2+2ab]$=-2\int e^t\left(\frac{\sqrt{(\cos t+\sin t{)}^2}}{2{\cos }^{2}t}\right)dt[\because (a+b{)}^2=a^2+b^2+2ab]$
    =−2∫et(cost+sint2cos2t)dt$=-2\int e^t\left(\frac{\cos t+\sin t}{2{\cos }^{2}t}\right)dt$
    =−∫et(sect+tantsect)dt$=-\int e^t(\sec t+\tan t\sec t)dt$[∵1cosx=secx,sinxcosx=tanx]$[\because \frac{1}{cosx}=secx,\frac{sinx}{cosx}=tanx]$
    we know that , ∫et[f(t)+f′(t)]dt=etf(t)+C$\int e^t\left[f(t)+f^{\mathrm{\prime }}(t)\right]dt=e^{t}f(t)+C$
    Now, consider f(t)=sec t$f(t)=sect$
    then f′(t)=sec t tan t$f^{\prime }(t)=secttant$
    ∴I=etsect+C$\therefore I=e^t\sec t+C$
    =−e−x/2secx2+C$=-e^{-x/2}\sec \frac{x}{2}+C$[∵t=−x2 and sec(−θ)=secθ]$\left[\because t=\frac{-x}{2}\text{ and }\sec (-\theta )=\sec \theta \right]$
    I=−e−x/2secx2+C$I=-e^{-x/2}\sec \frac{x}{2}+C$
17. I=∫01xlog(1+2x)dx$I=\underset{0}{\overset{1}{\int }}x\log \left(1+2x\right)dx$
    =[log(1+2x)x22]10−∫11+2x.2.x22dx$={\left[\log \left(1+2x\right)\frac{x^2}{2}\right]}_0^1-\int \frac{1}{1+2x}.2.\frac{x^2}{2}dx$
    =12[x2log(1+2x)]10−∫x21+2xdx$=\frac{1}{2}{\left[x^2\log \left(1+2x\right)\right]}_0^1-\int \frac{x^2}{1+2x}dx$
    =12[log3−0]10−[∫10(x2−x21+2x)dx]$=\frac{1}{2}{\left[\log 3-0\right]}_0^1-\left[{\int }_0^1\left(\frac{x}{2}-\frac{\frac{x}{2}}{1+2x}\right)dx\right]$
    =12log3−12∫10xdx+12∫10x1+2xdx$=\frac{1}{2}\log 3-\frac{1}{2}{\int }_0^{1}xdx+\frac{1}{2}{\int }_0^1\frac{x}{1+2x}dx$
    =12log3−12[x22]10+12∫1012(2x+1−1)(2x+1)dx$=\frac{1}{2}\log 3-\frac{1}{2}{\left[\frac{x^2}{2}\right]}_0^1+\frac{1}{2}{\int }_0^1\frac{\frac{1}{2}\left(2x+1-1\right)}{\left(2x+1\right)}dx$
    =12log3−12[12−0]$=\frac{1}{2}\log 3-\frac{1}{2}\left[\frac{1}{2}-0\right]$ +14∫10dx−14∫1011+2xdx$+\frac{1}{4}{\int }_0^{1}dx-\frac{1}{4}{\int }_0^1\frac{1}{1+2x}dx$
    =12log3−14+14[x]10−18[log|(1+2x)|]10$=\frac{1}{2}\log 3-\frac{1}{4}+\frac{1}{4}{\left[x\right]}_0^1-\frac{1}{8}{\left[\log \left|\left(1+2x\right)\right|\right]}_0^1$
    =12log3−14+14−18[log3−log1]$=\frac{1}{2}\log 3-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}\left[\log 3-\log 1\right]$
    =12log3−18log3$=\frac{1}{2}\log 3-\frac{1}{8}\log 3$
    =38log3$=\frac{3}{8}\log 3$
18. According to the question , I=∫31(2x2+5x)dx$I={\int }_1^3\left(2x^2+5x\right)dx$
    On comparing the given integral with ∫baf(x)dx,${\int }_a^{b}f(x)dx,$ we get
    a = 1, b = 3 and f(x) =2x2 + 5x
    We know that , ∫baf(x)dx=limh→0h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n−1)h) ]${\int }_a^{b}f(x)dx=\underset{h\to 0}{lim}h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]$...(i)
    where, h=b−an⇒nh=b−a  =3−1=2$$\begin{gathered} h=\frac{b-a}{n}\Rightarrow nh=b-a \\ =3-1=2 \end{gathered}$$
    f(a) = f(1) = 2(1)2 + 5(1) = 2 + 5 = 7
    f(a+h) = f(1 + h) = 2(1 + h)2 + 5(1 + h) = 2 + 2h2 + 4h + 5 + 5h = 2h2+ 9h + 7
    f(a + 2h) = 2(1 + 2h)2 + 5(1 + 2h) = 2 + 8h2 + 8h + 5 +10h = 8h2 + 18h + 7
    so on
    f(a+(n-1)h) = f{1 + (n - 1)h} = 2{1 + (n - 1)h}2 + 5{1 + (n - 1)h}2 = 2 + 2(n - 1)2h2 + 4(n - 1)h + 5 + 5(n - 1)h = 2(n - 1)2h2 + 9(n - 1)h + 7
    On putting all above values in (i), we get
    ∫31(2x2+5x)dx=limh→0${\int }_1^3\left(2x^2+5x\right)dx=\underset{h\to 0}{lim}$h[7 + (2h2 + 9h + 7) +(8h2 + 18h + 7).......+2(n - 1)2h2 + 9(n - 1)h + 7]
    On rearranging terms , we get
    =limh→0$=\underset{h\to 0}{lim}$h[7 + 7+ 7........+ 7] + limh→0$\underset{h\to 0}{lim}$h[2h2 + 8h2+.........+2(n - 1)2h2] + limh→0$\underset{h\to 0}{lim}$h[9h + 18h +......+9(n -1 )h]
    =limh→0$=\underset{h\to 0}{lim}$7nh + limh→0$\underset{h\to 0}{lim}$2h3[ 12 + 22 + 32 +.........+(n -1)2] +limh→0$\underset{h\to 0}{lim}$9h2[ 1+ 2 +........+(n - 1)]
    =limh→07(2)+limh→02h3⋅n(n−1)(2n−1)6$=\underset{h\to 0}{lim}7(2)+\underset{h\to 0}{lim}\frac{2h^3\cdot n(n-1)(2n-1)}{6}$+limh→09h2⋅n(n−1)2$+\underset{h\to 0}{lim}\frac{9h^2\cdot n(n-1)}{2}$[∵∑n=n(n+1)2,∑n2=n(n+1)(2n+1)6]$[\because \sum n=\frac{n(n+1)}{2},\sum n^2=\frac{n(n+1)(2n+1)}{6}]$
    =14+limh→0nh(nh−h)(2nh−h)3$=14+\underset{h\to 0}{lim}\frac{nh(nh-h)(2nh-h)}{3}$+ limh→092⋅nh(nh−h)$\underset{h\to 0}{lim}\frac{9}{2}\cdot nh(nh-h)$
    =14+2(2−0)(4−0)3+92⋅2(2−0)$=14+\frac{2(2-0)(4-0)}{3}+\frac{9}{2}\cdot 2(2-0)$
    =14+163+18$=14+\frac{16}{3}+18$
    =42+16+543$=\frac{42+16+54}{3}$
    =1123$=\frac{112}{3}$ sq units.