Definite Integration

The value of \(\int\limits_0^1 {{{8\log \left( {1 + x} \right)} \over {1 + {x^2}}}} dx\) is

Let \(p(x)\) be a function defined on \(R\) such that \(p'(x)=p'(1-x),\) for all \(x \in \left[ {0,1} \right],p\left( 0 \right) = 1\) and \(p(1)=41.\) Then \(\int\limits_0^1 {p\left( x \right)dx}\) equals :

\(\int\limits_0^\pi {\left[ {\cot x} \right]dx,}\) where \(\left[ . \right]\) denotes the greatest integer function, is equal to:

Let \(I = \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx}\) and \(J = \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx} .\) Then which one of the following is true?

Let \(F\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),\) where \(f\left( x \right) = \int\limits_l^x {{{\log t} \over {1 + t}}dt,}\) Then \(F(e)\) equals

The solution for \(x\) of the equation \(\int\limits_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }} = {\pi \over 2}}\) is

\(\int\limits_0^\pi {xf\left( {\sin x} \right)dx}\) is equal to

\(\int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} dx\) is equal to

The value of \(\int\limits_1^a {\left[ x \right]} f'\left( x \right)dx,a > 1\) where \({\left[ x \right]}\) denotes the greatest integer not exceeding \(x\) is

\(\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}}.... + {1 \over n}{{\sec }^2}1} \right]\)
equals

The value of \(\int\limits_{ - \pi }^\pi {{{{{\cos }^2}} \over {1 + {a^x}}}dx,\,\,a > 0,}\) is

Let \(f:R \to R\) be a differentiable function having \(f\left( 2 \right) = 6\),
\(f'\left( 2 \right) = \left( {{1 \over {48}}} \right)\). Then $$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2...

The value of integral, \(\int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx\) is

If \({I_1} = \int\limits_0^1 {{2^{{x^2}}}dx,{I_2} = \int\limits_0^1 {{2^{{x^3}}}dx,\,{I_3} = \int\limits_1^2 {{2^{{x^2}}}dx} } }\) and \({I_4} = \int\limits_1^2 {{2^{{x^3}}}dx}\) then

\(\mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{e^{{r \over n}}}}\) is

The value of \(\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|dx}\) is

If \(f\left( x \right) = {{{e^x}} \over {1 + {e^x}}},{I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}dx}\)
and $${I_2} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)...

If \(\int\limits_0^\pi {xf\left( {\sin x} \right)dx = A\int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx,} }\) then \(A\) is

The value of \(I = \int\limits_0^{\pi /2} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}dx}\) is

\(\mathop {\lim }\limits_{n \to \infty } {{1 + {2^4} + {3^4} + .... + {n^4}} \over {{n^5}}}\) - \(\mathop {\lim }\limits_{n \to \infty } {{1 + {2^3} + {3^3} + .... + {n^3}} \over {{n^5}}}\)

If \(f\left( y \right) = {e^y},\) \(g\left( y \right) = y;y > 0\) and
\(F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,}\) then :

If \(f\left( {a + b - x} \right) = f\left( x \right)\) then \(\int\limits_a^b {xf\left( x \right)dx}\) is equal to

The value of \(\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {{{\sec }^2}tdt} } \over xsinx}\) is

The value of the integral \(I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx}\) is

Let \(f(x)\) be a function satisfying \(f'(x)=f(x)\) with \(f(0)=1\) and \(g(x)\) be a function that satisfies \(f\left( x \right) + g\left( x \right) = {x^2}\). Then the value of the integral $$\int\limits_0^1 {f\left( x \right)g\left( x \...

\({I_n} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx}\) then \(\,\mathop {\lim }\limits_{n \to \infty } \,n\left[ {{I_n} + {I_{n + 2}}} \right]\) equals

\(\int\limits_0^{10\pi } {\left| {\sin x} \right|dx}\) is

\(\int_{ - \pi }^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} dx\) is

If \(y=f(x)\) makes +\(ve\) intercept of \(2\) and \(0\) unit on \(x\) and \(y\) axes and encloses an area of \(3/4\) square unit with the axes then \(\int\limits_0^2 {xf'\left( x \right)dx}\) is

\(\int\limits_0^2 {\left[ {{x^2}} \right]dx}\) is

\(\mathop {\lim }\limits_{n \to \infty } {{{1^p} + {2^p} + {3^p} + ..... + {n^p}} \over {{n^{p + 1}}}}\) is