Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^2 x\sqrt{2 - x} d x\]
\[ = \int_0^2 \left( 2 - x \right)\sqrt{2 - 2 + x} d x\]
\[ = \int_0^2 \left( 2 - x \right)\sqrt{x} d x\]
\[ = \int_0^2 \left( 2\sqrt{x} - x\sqrt{x} \right) dx\]
\[ = \int_0^2 \left( 2 x^\frac{1}{2} - x^\frac{3}{2} \right) dx\]
\[ = \left[ 2\frac{x^\frac{3}{2}}{\frac{3}{2}} - \frac{x^\frac{5}{2}}{\frac{5}{2}} \right]_0^2 \]
\[ = \left[ \frac{4}{3} x^\frac{3}{2} - \frac{2}{5} x^\frac{5}{2} \right]_0^2 \]
\[ = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5} \]
`=(5xx8sqrt2)/(3xx5)-(3xx8sqrt2)/(5xx3)`
`=(16sqrt2)/15`
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
`Γ (9/2)`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
