Advertisements
Advertisements
प्रश्न
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
Advertisements
उत्तर
\[Let, I = \int_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} d x ...................(1)\]
\[ = \int_2^3 \frac{\sqrt{5 - x}}{\sqrt{5 - 5 + x} + \sqrt{5 - x}} d x \]
\[ = \int_2^3 \frac{\sqrt{5 - x}}{\sqrt{x} + \sqrt{5 - x}} d x ...................(2)\]
Adding (1) and (2)
\[ 2I = \int_2^3 \left[ \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} + \frac{\sqrt{5 - x}}{\sqrt{x} + \sqrt{5 - x}} \right] d x\]
\[ = \int_2^3 \frac{\sqrt{5 - x} + \sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[ = \int_2^3 dx \]
\[ = \left[ x \right]_2^3 \]
\[ = 3 - 1 = 1\]
\[Hence, I = \frac{1}{2}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate the following integral:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
