Integral Calculus MCQ -

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Integral Calculus MCQ

1 Integrate x¹¹ with respect to x.

(a) x¹¹+ C

(b) X¹¹/12 + 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^–¹¹with respect to x.

(a) x-⁹ / – 9+ C

(b) X^-10/ -10 + 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) x^1-n/(1-n) + 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

= {(x^1-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

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

(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) x^2/5 +C

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

(d) None of the above

_{x√x dx =} _x¹_.x^1/2.dx = x^3/2.dx = { x^3/2+1 / ( $\displaystyle \frac{3}{2}$ +1) } +C

= ( x^5/2 / $\displaystyle \frac{5}{2}$ ) +C = $\displaystyle \frac{2}{5}$ x^5/2+ 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 $\displaystyle \int{{}}$ _e^mx _{dx = e}^mx / 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 / log_e5

(b) 5²/ log_e5

(c ) 5^2x/ 2log_e5

(d) 1

We know, $\displaystyle \int{{}}$ a^mx dx = a^mx/ (mlog_ea) +C.

So, $\displaystyle \int{{}}$ 5^2x dx = 5^2x / (2log_e5) +C

So, option (c) is correct

9. Integrate $\displaystyle \int{{}}$ {(e^4x+ e^2x ) / e^2x } dx
(a)       e^x + e^{– x}
(b)      e^x – e^{– x}
(c)       e^2x + e^{4 x}
(d)      (e^2x / 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 $\displaystyle \int{{}}$ _e^mx _{dx = e}^mx / m (m ¹ 0)]

So, option (d) is correct

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

(a) https://dvidya.com/integral-calculus-mcq/(1-x)⁴ +C

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

(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) $\displaystyle \frac{1}{x}$

(d) log x

Let us logx as 1^st 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

Integral Calculus MCQ

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