Advertisements
Advertisements
Question
Advertisements
Solution
\[\int_0^1 \frac{1}{1 + x^2} d x\]
\[ = \left[ \tan^{- 1} x \right]_0^1 \]
\[ = \tan^{- 1} 1 - \tan^{- 1} 0\]
\[ = \frac{\pi}{4} - 0\]
\[ = \frac{\pi}{4}\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
If f(2a − x) = −f(x), prove that
Write the coefficient a, b, c of which the value of the integral
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
