Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^1 x \left( 1 - x \right)^5 d x . Then, \]
\[I = \int_0^1 \left( x - 1 + 1 \right) \left( 1 - x \right)^5 d x\]
\[ \Rightarrow I = \int_0^1 \left[ - \left( 1 - x \right)^6 + \left( 1 - x \right)^5 \right] d x\]
\[ \Rightarrow I = \left[ \frac{\left( 1 - x \right)^7}{7} \right]_0^1 - \left[ \frac{\left( 1 - x \right)^6}{6} \right]_0^1 \]
\[ \Rightarrow I = - \frac{1}{7} + \frac{1}{6}\]
\[ \Rightarrow I = \frac{1}{42}\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
Evaluate each of the following integral:
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Find: `int logx/(1 + log x)^2 dx`
Evaluate \[\int_{1}^{3}(2x+1)\,dx\].
