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12th Business Maths Weekly Test - 4 ( Integral Calculus – I )

St. Britto Hr. Sec. School - Madurai

12th Business Maths Weekly Test - 4 ( Integral Calculus – I )-Aug 2020

Time : 1.00 Hours

Part - A

Multiple Choice Question - Answer All Question

The value of \(\int _{ 2 }^{ 3 }{ f(5-x) } dx-\int _{ 2 }^{ 3 }{ f(x) } dx\) is

1
0
-1
5

ഽ\(\frac { { 2x }^{ 3 } }{ 4+{ x }^{ 4 } } \)dx is

\(log\left| 4+{ x }^{ 4 } \right| +c\)
\(\frac { 1 }{ 2 } log\left| 4+{ x }^{ 4 } \right| +c\)
\(\frac { 1 }{4 } log\left| 4+{ x }^{ 4 } \right| +c\)
\(log\left| \frac { { 2x }^{ 3 } }{ { 4+x }^{ 4 } } \right| +c\)

\(\int _{ 0 }^{ \frac { \pi }{ 3 } }{ tanx } dx\) is

log 2
0
log\(\sqrt { 2 } \)
2 log 2

ഽ\(\frac{logx}{x}\)dx , x > 0 is

\(\frac12\)(log x)² + c
-\(\frac12\)(log x)²
\(\frac { 2 }{ { x }^{ 2 } } +c\)
\(\frac { 2 }{ { x }^{ 2 } } +c\)

The value of \(\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ cosx } \) dx is

0
2
1
4

ഽ\(\frac { 1 }{ { x }^{ 3 } } \)dx is

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

\(\int _{ 0 }^{ \infty }{ { x }^{ 4 }{ e }^{ -x } } \)dx is

12
4
4!
64

ഽ\(\frac { dx }{ \sqrt { { x }^{ 2 }-{ 36 } } } \) is

\(\sqrt { { x }^{ 2 }-{ 36 } } +c\)
log\(\left| x+\sqrt { { x }^{ 2 }-36 } \right| +c\)
log\(\left| x-\sqrt { { x }^{ 2 }-36 } \right| +c\)
\(log\left| { x }^{ 2 }+\sqrt { { x }^{ 2 }-36 } \right| +c\)

ഽ2^xdx is

2^x log 2 + c
2^x+ c
\(\frac { 2^{ x } }{ log2 } +c\)
\(\frac { log2 }{ { 2 }^{ x } } +c\)

If \(\int _{ 0 }^{ 1 }{ f(x) } dx=1,\int _{ 0 }^{ 1 }{ xf(x) } dx=a\) and \(\int _{ 0 }^{ 1 }{ { x }^{ 2 }f(x) } dx={ a }^{ 2 }\), then \(\int _{ 0 }^{ 1 }{ { (a-x) }^{ 2 } } f(x)\) is

4a²
0
2a²
1

\(\int _{ -1 }^{ 1 }{ { x }^{ 3 }{ e }^{ { x }^{ 4 } } } \) dx is

1
2\(\int _{ -1 }^{ 1 }{ { x }^{ 3 }{ e }^{ { x }^{ 4 } } } \)dx
0
\({ e }^{ { x }^{ 4 } }\)

\(\int _{ 0 }^{ 4 }{ \left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) } \)dx is

\(\frac{20}{3}\)
\(\frac{21}{3}\)
\(\frac{28}{3}\)
\(\frac{1}{3}\)

\(\int _{ 0 }^{ 1 }{ (2x+1) } dx\) is

1
2
3
4

\(\Gamma (n)\) is

(n −1)!
n!
\(n\Gamma (n)\)
(n −1)\(\Gamma \)(n)

ഽ\(\frac { { e }^{ x } }{ \sqrt { { 1+e }^{ x } } } \)dx is

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

If f (x) is a continuous function and a < c < b ,then \(\int _{ a }^{ c }{ f(x) } dx+\int _{ c }^{ b }{ f(x) } dx\) is

\(\int _{ a }^{ b }{ f(x) } dx+\int _{ a }^{ c }{ f(x) } dx\)
\(\int _{ a }^{ c }{ f(x) } dx+\int _{ a }^{ b }{ f(x) } dx\)
\(\int _{ a }^{ b }{ f(x) } dx\)
0

ഽ\(\frac { sin5x-sinx }{ cos3x } \)dx

−cos2x + c
−cos2x + c
\(-\frac14\)cos2x c
−4cos2x + c

ഽ\(\frac { sin2x }{ 2sinx } dx\) is

sin x + c
\(\frac12\)sin x + c
cos x + c
\(\frac12\)cos x + c

ഽ\(\frac { { e }^{ x } }{ { e }^{ x }+1 } \) dx

log\(\left| \frac { { e }^{ x } }{ { e }^{ x }+1 } \right| +c\)
log\(\left| \frac { { e }^{ x }+1 }{ { e }^{ x } } \right| +c\)
log\(\left| { e }^{ x } \right| +c\)
log\(\left| { e }^{ x }+1 \right| +c\)

\(\int _{ 0 }^{ 1 }{ \sqrt { { x }^{ 4 }({ 1-x) }^{ 2 } } } dx\) is

\(\frac{1}{12}\)
\(\frac{-7}{12}\)
\(\frac{7}{12}\)
\(\frac{-1}{12}\)

ഽ\(\sqrt { { e }^{ x } } \) dx is

\(\sqrt { { e }^{ x } } +c\)
\(2\sqrt { { e }^{ x } } \) + c
\(\frac12\sqrt { { e }^{ x } } +c\)
\(\frac { 1 }{ 2\sqrt { { e }^{ x } } } +c\)

ഽe^2x[2x² + 2x]dx

e^2xx² + c
xe^2x + c
2x²e² + c
\(\frac { { x }^{ 2 }{ e }^{ x } }{ 2 } +c\)

ഽ\(\frac { 2x+3 }{ \sqrt { x^{ 2 }+3x+2 } } \) dx is

\(\sqrt { x^{ 2 }+3x+2 } \) + c
\(2\sqrt { x^{ 2 }+3x+2 } +c\)
log(x² + 3x + 2)+ c
\(\frac23\)(x²+ 3x + 2)^{\(\frac { 3 }{ 2 } \)}

\(\Gamma (1)\) is

0
1
n
n!

Using the factorial representation of the gamma function, which of the following is the solution for the gamma function \(\Gamma \)(n) when n = 8

5040
5400
4500
5540

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