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11th Maths Weekly Test - 3( Integral Calculus )

St. Britto Hr. Sec. School - Madurai

11th Maths Weekly Test - 3( Integral Calculus )-Aug 2020

Time : 1.00 Hours

Part - A

Multiple Choice Question - Answer All Question

\(\int { \frac { 1 }{ 2 } } { sec }^{ 2 }\) x dx is _____ + c.

\(\frac { 1 }{ 2 } \) tan x
tan x
2 tan x
none of these

\(\int x^2cos \ x \ dx\) is

x² sin x+2x cos x-2sin x+c
x² sin x-2x cos x-2sin x+c
-x² sin x+2x cos x-2sin x+c
-x² sin x-2x cos x-2sin x+c

\(\int {dx\over e^x-1}\)is

\(log|e^x|-log|e^x-1|+c\)
\(log|e^x|+log|e^x-1|+c\)
\(log|e^x-1|-log|e^x|+c\)
\(log|e^x+1|-log|e^x|+c\)

\(\int { \frac { sin\sqrt { x } }{ x } } \) dx = ________ +c.

2 cos \(\sqrt { x } \)
2 sin \(\sqrt { x } \)
-2 sin \(\sqrt { x } \)
-2 cos \(\sqrt { x } \)

\(\int {e^x(1+x)\over cos^2(xe^x)}dx\) is

cot(xe^x)+c
sec(xe^x)+c
tan(xe^x)+c
cos(xe^x)+c

Match List - I with List II.

List - I List - II

i \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log(tan\quad x)dx } \) a \(\frac { 16 }{ 35 } \)

ii \(\int _{ 0 }^{ 1 }{ x{ (1-x) }^{ 10 }dx } \) b \(\frac { 120 }{ 46 } \)

iii \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ si{ n }^{ 7 }xdx } \) c \(\frac { 1 }{ 132 } \)

iv \(\int _{ 0 }^{ \infty }{ { x }^{ 5 }{ e }^{ -4x }dx } \) d 0

The Correct match is

i ii iii iv

d c a b
i ii iii iv

d c b a
i ii iii iv

b d c a
i ii iii iv

b c a d

\(\int e^{\sqrt{x}}dx\) is

\(2\sqrt{x}(1-e^{\sqrt{x}})+c\)
\(2\sqrt{x}(e^{\sqrt{x}}-1)+c\)
\(2e^{\sqrt{x}}(1-\sqrt{x})+c\)
\(2e^{\sqrt{x}}(\sqrt{x}-1)+c\)

The gradient (slope) of a curve at any point (x, y) is\({x^2-4\over x^2}\). If the curve passes through the point (2, 7), then the equation of the curve is

\(y=x+{4\over x}+3\)
\(y=x+{4\over x}+4\)
y=x²+3x+4
y=x²-3x+6

\(\int \sqrt{{1-x\over 1+x}}dx\) is

\(\sqrt{1+x^2}+sin^{-1}x+c\)
\(sin^{-1}x-\sqrt{1+x^2}+c\)
\(log|x+\sqrt{1-x^2}|-\sqrt{1-x^2}+c\)
\(\sqrt{1-x^2}+log|x+\sqrt{1-x^2}|+c\)

\(\int 2^{3x+5}dx\) is

\({3(2^{3x+5})\over log 2}+c\)
\({2^{3x+5}\over log 2(3x+5)}+c\)
\({2^{3x+5}\over 2log 3}+c\)
\({2^{3x+5}\over 3log 2}+c\)

\(\int { \frac { 4\left( sin^{ -1 }x \right) ^{ 3 } }{ \sqrt { 1-{ x }^{ 2 } } } } \) dx = ________+c.

log (sin ^-1x)
(sin ^-1 x)⁴
4 (sin^-1 x)⁴
\(\frac { \left( { sin }^{ -1 }x \right) ^{ 4 } }{ 4 } \)

\(\int { { e }^{ x }\left[ f\left( x \right) +f'\left( x \right) \right] } \) dx = ___________+c.

e^x f (x)
e^x f (x)
2e^x f(x)
e^x - f (x)

\(\int { \frac { \left( log{ x } \right) ^{ 3 } }{ x } } \) dx = _________+c.

\(\frac { \left( log{ x } \right) ^{ 4 } }{ 4 } \)
(sin^-1 x)⁴
(log x)⁴
\(\frac { 1 }{ 3logx } \)

\(\int {sec \ x\over \sqrt{cos 2x}}dx\) is

tan^-1(sin x)+c
2sin^-1(tan x)+c
tan^-1(cos x)+c
sin ^-1(tan x)+c

\(\int{x^2+cos^2x\over x^2+1}cosec^2xdx\) is

cot x+sin ^-1x+c
-cot x+tan^-1x+c
-tan x+cot^-1x+c
-cot x-tan^-1x+c

\(\int { \frac { x }{ 4+{ x }^{ 4 } } } \) dx is equal to________+c.

\(\frac { 1 }{ 4 } \) tan^-1 (x²)
\(\frac { 1 }{ 4 } \) tan^-1 \(\left( \frac { { x }^{ 2 } }{ 2 } \right) \)
\(\frac { 1 }{ 2 } \) tan^-1 \(\left( \frac { { x }^{ 2 } }{ 2 } \right) \)
none of these

If \(\int {3^{1\over x}\over x^2}dx=k(3^{1\over x})+c\) ,then the value of k is

log 3
-log 3
\(-{1\over log3}\)
\({1\over log3}\)

If\(\int f(x)dx=g(x)+c\) ,then\(\int f(x)g'(x)dx\)

\(\int (f(x))^2dx\)
\(\int f(x)g(x)dx\)
\(\int f'(x)g(x)dx\)
\(\int (g(x))^2dx\)

\(\int {1\over x\sqrt{(log \ x)^2+5}}dx\)is

\(log|x+\sqrt{x^2-5}|+c\)
\(log|logx+\sqrt{logx-5}|+c\)
\(log|logx+\sqrt{(logx)^2-5}|+c\)
\(log|logx-\sqrt{(logx)^2-5}|+c\)

\(\int {sin^8x-cos^8x\over 1-2sin^2 \ x \ cos^2 \ x}dx\) is

\({1\over2}sin 2x+c\)
\(-{1\over2}sin 2x+c\)
\({1\over2}cos 2x+c\)
\(-{1\over2}cos 2x+c\)

\(\int {x+2\over \sqrt{x^2-1}}dx\)is

\(\sqrt{x^2-1}-2 log|x+\sqrt{x^2-1}|+c\)
\(sin^{-1}x-2 log|x+\sqrt{x^2-1}|+c\)
\(2 log|x+\sqrt{x^2-1}|-sin^{-1}x+c\)
\(\sqrt{x^2-1}+2log|x+\sqrt{x^2-1}|+c\)

\(\int { \frac { 1 }{ 9x^{ 2 }-4 } } \) dx = ____________+c.

log\(\left| \frac { 3x-2 }{ 3x+2 } \right| \)
\(\frac { 1 }{ 12 } \)log\(\left| \frac { 3x-2 }{ 3x+2 } \right| \)
12log\(\left| \frac { 3x-2 }{ 3x+2 } \right| \)
\(\frac { 1 }{ 12 } log\left| \frac { 3x+2 }{ 3x-2 } \right| \)

\(\int{ {e^x}(x^2 \ tan^{-1}x+tan^{-1}x+1)\over x^2+1}dx\) is

e^xtan^-1(x+1)+c
tan^-1(e^x)+c
\(e^x{(tan^{-1}x)^2\over 2}+c\)
e^x tan^-1 x+c

\(\int { { tan }^{ 3 } } 2sec2x\) dx = __________________+c.

\(\frac { 1 }{ 6 } \) sec³ 2x
\(\frac { 1 }{ 6 } \) sec³2x - \(\frac { 1 }{ 2 } \) sec 2x
\(\frac { 1 }{ 2 } \) sec 2x
\(\frac { 1 }{ 6 } \) sec³ 2x + \(\frac { 1 }{ 2 } \) sec 2x

\(\int { \frac { \left( 1+logx \right) ^{ 2 } }{ x } } \) dx = _______+C.

\(\frac { \left( 1+logx \right) ^{ 3 } }{ 3 } \)
3 log ( 1 + log x)
2(1 + log x)
none of these

\(\int sin^3 \ xdx \) is

\({-3\over 4}cos \ x-{cos \ 3x\over 12}+c\)
\({3\over 4}cos \ x-{cos \ 3x\over 12}+c\)
\({-3\over 4}cos \ x+{cos \ 3x\over 12}+c\)
\({-3\over 4}sin \ x-{sin \ 3x\over 12}+c\)

\(\int {e^{6logx}-e^{5logx}\over e^{4logx}-e^{3logx}}dx\) is

x+c
\({x^3\over 3}+c\)
\({3\over x^3}+c\)
\({1\over x^2}+c\)

\(\int tan^{-1}\sqrt{1-cos \ 2x\over 1+cos \ 2x}dx\) is

x²+c
2x²+c
\({x^2\over2}+c\)
\(-{x^2\over2}+c\)

\(\int e^{-4x}cos \ x \ d x\) is

\({e^{-4x}\over 17}[4cos \ x-sin \ x]+c\)
\({e^{-4x}\over 17}[-4cos \ x+sin \ x]+c\)
\({e^{-4x}\over 17}[4cos \ x+sin \ x]+c\)
\({e^{-4x}\over 17}[-4cos \ x-sin \ x]+c\)

\(\int x^2e^{x\over2}dx\) is

\( x^2e^{x\over2}-4xe^{x\over2}-8e^{x\over2}+c\)
\( 2x^2e^{x\over2}-8xe^{x\over2}-16e^{x\over2}+c\)
\( 2x^2e^{x\over2}-8xe^{x\over2}+16e^{x\over2}+c\)
\( x^2{e^{x\over2}\over 2}-{xe^{x\over2}\over 4}+{e^{x\over2}\over 8}+c\)

If\(\int f'(x)e^{x^2}dx=(x-1)e^{x^2}+c\),then f(x) is

\(2x^3-{x^2\over 2}+x+c\)
\({x^3\over 2}+3x^2+4x+c\)
x³+4x²+6x+c
\({2x^3\over3}-x^3+x+c\)

\(\int { sin } \)e^x . d (e^x) = ________+c.

cos (e^x)
sin(e^x)
- cos (e^x)
-sin(e^x)

\(\int {\sqrt{tan \ x}\over sin \ 2x}dx\) is

\(\sqrt{tan \ x}+c\)
\(2\sqrt{tan \ x}+c\)
\({1\over2}\sqrt{tan \ x}+c\)
\({1\over4}\sqrt{tan \ x}+c\)

\(\int { { 3 }^{ x+2 } } \) dx = __________+c.

\(\frac { { 3 }^{ x } }{ log3 } \)
\(9\left( \frac { { 3 }^{ x } }{ log3 } \right) \)
\(\frac { { 3 }^{ x } }{ 9log3 } \)
3^x.9

\(\int { x } \) sin x dx = -x cos x + a, then a =

sin x + c
cos x + c
c
none of these

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