Advertisements
Advertisements
Question
Options
π/4
π/2
π
1
Advertisements
Solution
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
RELATED QUESTIONS
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
