Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Advertisements
उत्तर
`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
APPEARS IN
संबंधित प्रश्न
If f(x) is a continuous function defined on [−a, a], then prove that
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
