Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let I = \int_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}} d x......................(1)\]
\[I = \int_0^5 \frac{\sqrt[4]{9 - x}}{\sqrt[4]{9 - x} - \sqrt[4]{x + 4}}dx ...........................\left[\text{Using }\int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]
\[I = - \int_0^5 \frac{\sqrt[4]{9 - x}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}}dx ...................(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}} - \frac{\sqrt[4]{9 - x}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}}dx \]
\[ = \int_0^5 \frac{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}}dx\]
\[ = \int_0^5 dx\]
\[ = \left[ x \right]_0^5 \]
\[ = 5\]
\[Hence\ I = \frac{5}{2}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
Write the coefficient a, b, c of which the value of the integral
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
Which integral has Constant C present?
Which integral has lower and upper limits?
Evaluate \[\int_{0}^{1}x\,dx\].
