if a 0 1 e t t 1dt then 0 1 e tlog 1 t dt in terms of a

If $$A = \int\limits_0^1 {\frac{{{e^t}}}{{t + 1}}dt,} $$ then $$\int\limits_0^1 {{e^t}\log \left( {1 + t} \right)} dt$$ in terms of $$A$$ equals :

A. $$e\,\log \left( A \right)$$

B. $$\frac{e}{2} - A$$

C. $$e - 1 - \frac{A}{2}$$

D. $$\frac{e}{2} - 1 - A$$

Answer : $$\frac{e}{2} - 1 - A$$

Solution :
$$\eqalign{ & \int\limits_0^1 {{e^t}\log \left( {1 + t} \right)} dt \cr & = \left[ {{e^t}.\frac{1}{{1 + t}}} \right]_0^1 - \int\limits_0^1 {\frac{{{e^t}}}{{t + 1}}dt} \cr & = \frac{e}{2} - 1 - A \cr} $$

Releted MCQ Question on
Calculus >> Definite Integration

Releted Question 1

The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-

A. $$ - 1$$

B. $$2$$

C. $$1 + {e^{ - 1}}$$

D. none of these

View Answer

Releted Question 2

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

A. no root in $$\left( {0,\,2} \right)$$

B. at least one root in $$\left( {0,\,2} \right)$$

C. a double root in $$\left( {0,\,2} \right)$$

D. two imaginary roots

View Answer

Releted Question 3

The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$ is-

A. $$\frac{\pi }{4}$$

B. $$\frac{\pi }{2}$$

C. $$\pi $$

D. none of these

View Answer

Releted Question 4

For any integer $$n$$ the integral $$\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^3}\left( {2n + 1} \right)xdx$$ has the value-

A. $$\pi $$

B. $$1$$

C. $$0$$

D. none of these

View Answer

If $$A = \int\limits_0^1 {\frac{{{e^t}}}{{t + 1}}dt,} $$ then $$\int\limits_0^1 {{e^t}\log \left( {1 + t} \right)} dt$$ in terms of $$A$$ equals :

Releted MCQ Question on
Calculus >> Definite Integration

The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$ is-

For any integer $$n$$ the integral $$\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^3}\left( {2n + 1} \right)xdx$$ has the value-

Practice More Releted MCQ Question on
Definite Integration

Practice More MCQ Question on Maths Section

If $$A = \int\limits_0^1 {\frac{{{e^t}}}{{t + 1}}dt,} $$ then $$\int\limits_0^1 {{e^t}\log \left( {1 + t} \right)} dt$$ in terms of $$A$$ equals :

Releted MCQ Question on Calculus >> Definite Integration

The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$ Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$ is-

For any integer $$n$$ the integral $$\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^3}\left( {2n + 1} \right)xdx$$ has the value-

Practice More Releted MCQ Question on Definite Integration

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Releted MCQ Question on
Calculus >> Definite Integration

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

Practice More Releted MCQ Question on
Definite Integration