Advertisements
Advertisements
Question
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Sum
Advertisements
Solution
`int_(-1)^1 "f"(x) "d"x = int_(-1)^0 "f"(x) "d"x + int_0^1 "f"(x) "d"x`
= `int_(-1)^0 (-x) "d"x + int_0^1 x "d"x`
= `- [x^2/2]_(-1)^0 + [x^2/2]_0^1`
= `- [0 - (-1)^2/2] + [(1)^2/2 - ((0))/2]`
= `- [-1/2] + [1/2]`
= `1/2 + 1/2`
= 1
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^4 \frac{1}{\sqrt{4x - x^2}} dx\]
\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]
\[\int\limits_0^1 \frac{1 - x^2}{x^4 + x^2 + 1} dx\]
\[\int\limits_0^2 x\sqrt{2 - x} dx\]
\[\int\limits_0^5 \left( x + 1 \right) dx\]
\[\int\limits_1^4 \left( x^2 - x \right) dx\]
Evaluate each of the following integral:
\[\int_0^\frac{\pi}{2} e^x \left( \sin x - \cos x \right)dx\]
\[\int\limits_0^2 x\left[ x \right] dx .\]
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
Find: `int logx/(1 + log x)^2 dx`
