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
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
Evaluate :
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
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
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
`int_0^(2a)f(x)dx`
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Choose the correct alternative:
Γ(1) is
