Integral Calculus MCQ

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Integral Calculus MCQ

1 Integrate x11 with respect to x.

(a)        x11 + C

(b)        X11/12 + C

(c)        X12/12 + C

(d)        x12 + C

We know  xn dx = (xn + 1) / (n+1).

So, x11 dx = {(x11 + 1) / (11+1)}  + C =  {(x12) / (12)} + C

So, option (c) is correct

2. Integrate x 11 with respect to x.

(a)        x-9 / – 9 + C

(b)        X-10 / -10 + C

(c)        X-11 / -11+ C

(d)        x12 + C

We know  xn dx = (xn + 1) / (n+1).

So, x-11 dx = {(x-11 + 1)/(-11+1)}  + C

=  {(x-10) / (-10)} + C

So, option (b) is correct

3. $\displaystyle Integrate\text{ }\frac{1}{{{{x}^{n}}}}\text{ with respect to x}\text{.}$

(a)        (x-n /- n) + C

(b)        x1-n/(1-n) + C

(c)        (x-n +1/n +1)+ C

(d)        1

We know  xn dx = (xn + 1) / (n+1).

1/ xn = x-n. So, x-n dx = {(x-n + 1) / (-n+1)}  + C

=  {(x1-n) / ( 1-n)} +C

So, option (b) is correct

4. $\displaystyle Integrate\text{ }\frac{1}{x}\text{ with respect to x}$

(a)           ax 1  + C

(b)           log (x) + C

(c)           Log (2) + C

(d)          $\displaystyle \frac{1}{x}$  + C

We know  $\displaystyle \frac{1}{x}$= x-1.  So, x-1  dx =$\displaystyle \int{{\frac{1}{x}}}$dx = log x +C

So, option (b) is correct

5. Integrate $\displaystyle 5\sqrt{x}~\text{ dx}$ with respect to x.

(a)           $\displaystyle \frac{5}{6}{{x}^{{6/5}}}+C$

(b)            $\displaystyle \frac{2}{5}{{x}^{{2/5}}}+C$

(c)            $\displaystyle \frac{2}{5}{{x}^{{5/2}}}+C$

(d)           None of the above

$\displaystyle \int{{5\sqrt{x}\text{ dx=5}\sqrt{x}}}$   dx 5 (x)1/2 dx =  5. (x) (1/2+1)  = 5. {(x) (1/2+1) } / ($\displaystyle \frac{1}{2}$ +1)

= 5. {(x) (3/2) / $\displaystyle \left( {\frac{3}{2}} \right)$ +C ={5. $\displaystyle \left( {\frac{2}{3}} \right)$. (x) (3/2)  +C = $\displaystyle {\frac{{10}}{3}}$ . x (3/2) + C

This value matches none of the options. So, option (d) is correct

6. Integrate $\displaystyle \int{{x\sqrt{x}\text{ dx}}}$.

(a)           x2/5 +C

(b)          $\displaystyle \frac{4}{5}$  x 2/5 +C

(c)            $\displaystyle \frac{2}{5}$ x 5/2 +C

(d)           None of the above

x√x dx  =  x1.x1/2 .dx =    x3/2.dx  = { x3/2+1 / ( $\displaystyle \frac{3}{2}$ +1) } +C

= ( x5/2 / $\displaystyle \frac{5}{2}$ )  +C =  $\displaystyle \frac{2}{5}$ x5/2 + C

So, option (c) is correct

7. Integrate $\displaystyle \int{{{{\text{e}}^{{-4x}}}\text{ dx}}}$

(a)       e 4x

(b)      4e 4x

(c )      e 4x  / ( – 4)

(d)      e 4x  / 4

We know   $\displaystyle \int{{}}$emx dx = emx / m (m $\displaystyle \ne$ 0).

So, $\displaystyle \int{{}}$ e– 4x dx = e– 4x / (-4) + C

So, option (c) is correct

8. Integrate $\displaystyle \int{{{{5}^{{2x}}}\text{ dx}}}$

(a)       5 x / loge5

(b)      5/ loge5

(c )      52x  / 2loge5

(d)      1

We know, $\displaystyle \int{{}}$amx dx = amx / (mlogea) +C.

So, $\displaystyle \int{{}}$52x dx = 52x  / (2loge5) +C

So, option (c) is correct

9. Integrate $\displaystyle \int{{}}${(e 4x+ e 2x ) / e 2x  } dx
(a)       ex + e– x
(b)      ex – e– x
(c)       e2x + e4 x
(d)      (e2x / 2) + x

$\displaystyle \int{{}}${(e 4x + e 2x ) / e 2x } dx

$\displaystyle \int{{}}$[{(e 4x / e 2x ) + (e 2x  / e 2x) }  dx

=   $\displaystyle \int{{}}$(e 4x-2x ) + (e 2x-2x) } dx

=  $\displaystyle \int{{}}${(e 4x-2x ) + 1} dx

=  $\displaystyle \int{{}}$ {(e 2x ) + 1 }  dx = (e 2x  / 2) + x

[as $\displaystyle \int{{}}$emx dx = emx / m (m ¹ 0)]

So, option (d) is correct

10. Evaluate $\displaystyle \int{{}}$ (1 – x)3 dx

(b)       $\displaystyle \frac{1}{4}$(1-x)4 +C

(c)     1 – x2

(d)     x2

Let 1 – x = t, So,  $\displaystyle \frac{{dt}}{{dx}}$(1-x) = -1.  Or,  $\displaystyle \frac{{dt}}{{dx}}$=-1, Or  dx= $\displaystyle \frac{{dt}}{{-1}}$

So, $\displaystyle \int{{}}$ (1 – x)3 dx = $\displaystyle \int{{}}$ t3 . $\displaystyle \frac{{dt}}{{-1}}$ = $\displaystyle \int{{}}$ t3. (– 1) dt = (-1) $\displaystyle \int{{}}$ t3. Dt

= (-1). (t3+1 / (3+1) = (-1). (t4 / 4)  + C

=  $\displaystyle -\frac{1}{4}$t4 +C = $\displaystyle -\frac{1}{4}$ (1-x)4 +C

So, option (a) is correct

11. Evaluate $\displaystyle \int{{}}$ log x dx

(a)       x log x – x + c

(b)     $\displaystyle \frac{1}{x}$

(c)       x log x – 1 + c

(d)      log x

Let us logx as 1st function and 1 as second function.

So, $\displaystyle \int{{}}$ log x dx = $\displaystyle \int{{}}$ (log x) . (1)  dx

= (log x) $\displaystyle \int{{}}$ (1) dx  –  $\displaystyle \int{{}}${$\displaystyle \frac{d}{{dx}}$ (log x) .  (1)} dx

= (log x) $\displaystyle \int{{}}$ (1) dx  –  $\displaystyle \int{{}}$ $\displaystyle \frac{d}{{dx}}$(log x) $\displaystyle \int{{}}$ (1) } dx

= x. (log x) – $\displaystyle \int{{}}$ $\displaystyle \frac{1}{x}$. x dx

= x. (log x) –$\displaystyle \int{{}}$ $\displaystyle \frac{1}{x}$. x dx = x. (log x) – $\displaystyle \int{{}}$dx

= x log x – x  +C   = x (log x – 1) +C

So, option (a) is correct

12. Evaluate $\displaystyle \int{{}}$ (5 + 2x)2 dx

(a) $\displaystyle \frac{1}{3}$(5 + 2x)3 + C
(b) $\displaystyle \frac{1}{6}$(5 + 2x)3 + C
(c) $\displaystyle \frac{1}{2}$(5 + 2x)3 + C
(d)  2x3+ C

Let 5 + 2x = t. So, $\displaystyle \frac{{dt}}{{dx}}$=$\displaystyle \frac{d}{{dx}}$ (5+2x) =2. So, dx=$\displaystyle \frac{{dt}}{2}$
So $\displaystyle \int{{}}$ (5 + 2x)2 dx = t2 + $\displaystyle \frac{1}{2}$dt =   $\displaystyle \frac{1}{2}$$\displaystyle \int{{}}$ t2.dt
= $\displaystyle \frac{1}{2}$ {(t((2+1) / (2+1)} + C =$\displaystyle \frac{1}{2}$  {(t3 / 3)} + C
=  $\displaystyle \frac{1}{6}$t3 + C =  $\displaystyle \frac{1}{6}$ (5 + 2x)3 + C.
So, option (b) is correct

13. Value of  is $\displaystyle \int\limits_{3}^{6}{{{{x}^{2}}dx}}\text{ is}$
(a) 46
(b) 210
(c) 63
(d) 72

$\displaystyle \int_{3}^{6}{{{{x}^{2}}}}dx$
$\displaystyle =\left[ {\frac{{{{x}^{{2+1}}}}}{{2+1}}} \right]_{3}^{6}$
$\displaystyle =\left[ {\frac{{{{x}^{3}}}}{3}} \right]_{3}^{6}$
$\displaystyle =\frac{{{{{(6)}}^{3}}}}{3}-\frac{{{{{\left( 3 \right)}}^{3}}}}{3}$
$\displaystyle =\frac{{216}}{3}-\frac{{27}}{3}$
= 72 – 9 = 63

14. Evaluate: $\displaystyle \int_{1}^{2}{{\log x\text{ dx}}}$
(a) log 2 – 1
(b) 2 log 2 + 1
(c) 2 log 2 – 1
(d) 1

$\displaystyle \int_{1}^{2}{{\log x\text{ dx}}}$
$\displaystyle I=\log x$
$\displaystyle {\left{ {\frac{d}{{dx}}(\log x)\int{{1.dx}}} \right}}$
$\displaystyle x\log x-\int{{\frac{1}{x}}}.xdx$
$\displaystyle x\log x-\int{{dx}}$
= x log x – x  = x (log x – 1)
$\displaystyle \therefore \int_{1}^{2}{{\log x\text{ dx}}}$
$\displaystyle =\int_{1}^{2}{{x(\log x-1)}}$
$\displaystyle =\left[ {x(\log x-1)} \right]_{1}^{2}$

= [2. (log 2 – 1)] – [1. (log 1 – 1)]
= 2 log 2 – 2 – [1. (0 – 1)]
= 2 log 2 – 2 – [1.(– 1)]
= 2 log 2 – 2 + 1 = 2 log 2 – 1
So, option (c) is correct

15. $\displaystyle Find\text{ }\int_{1}^{2}{{\log x\text{ dx}}}$
(a)$\displaystyle \frac{{{{e}^{4}}}}{2}$
(b) $\displaystyle {{{e}^{2}}}$
(c) $\displaystyle {{{e}^{4}}}$
(d) $\displaystyle \frac{1}{2}\left( {{{e}^{4}}-{{e}^{2}}} \right)$

$\displaystyle \int_{1}^{2}{{{{e}^{{2x}}}\text{ dx}}}$
$\displaystyle \int{{{{e}^{{2x}}}\text{ dx}=\frac{{{{e}^{{2x}}}}}{2}}}\left[ {\int{{{{e}^{{mx}}}dx\frac{{{{e}^{{mx}}}}}{m}}}} \right]$
Now $\displaystyle \int_{1}^{2}{{{{e}^{{2x}}}\text{ dx}}}$
$\displaystyle =\left[ {\frac{{{{e}^{{2x}}}}}{2}} \right]_{1}^{2}$
$\displaystyle =\left[ {\frac{{{{e}^{{22}}}}}{2}} \right]-\left[ {\frac{{{{e}^{{21}}}}}{2}} \right]$
$\displaystyle =\frac{{{{e}^{4}}}}{2}-\frac{{{{e}^{2}}}}{2}$
$\displaystyle =\frac{1}{2}\left( {{{e}^{4}}-{{e}^{2}}} \right)$
So, option (d) is correct