Advertisements
Advertisements
Question
Options
1
2
− 1
− 2
Solution
2
\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin\left| x \right| d x\]
\[ = - \int_{- \frac{\pi}{2}}^0 \sin x\ dx + \int_0^\frac{\pi}{2} \sin x\ dx\]
\[ = - \left[ - \cos x \right]_{- \frac{\pi}{2}}^0 + \left[ - \cos x \right]_0^\frac{\pi}{2} \]
\[ = 1 - 0 - 0 + 1\]
\[ = 2\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
Evaluate the following integral:
Evaluate each of the following integral:
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 \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
`int_0^(2a)f(x)dx`
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Choose the correct alternative:
Γ(1) is
Choose the correct alternative:
If n > 0, then Γ(n) is
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`