Advertisements
Advertisements
प्रश्न
पर्याय
π/4
π/2
π
1
Advertisements
उत्तर
1
\[\text{We have}, \]
\[ I = \int_0^\frac{\pi}{2} x \sin x\ d x \]
\[ = \left[ - x \cos x \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 1\left( - \cos x \right) d x\]
\[ = \left[ - x \cos x \right]_0^\frac{\pi}{2} + \int_0^\frac{\pi}{2} \cos x\ d x\]
\[ = - \left[ x \cos x \right]_0^\frac{\pi}{2} + \left[ \sin x \right]_0^\frac{\pi}{2} \]
\[ = - \left[ 0 - 0 \right] + \left[ 1 - 0 \right]\]
\[ = 1\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
If f is an integrable function, show that
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
If velocity changes with time, what does a definite integral give over a time interval?
