Advertisements
Advertisements
Question
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Sum
Advertisements
Solution
`int_0^1 "e"^(2x) "d"x = ["e"^(2x)/2]_0^1`
= `1/2 ["e"^(2x)]_0^1`
= `1/2["e"^(2(1)) - "e"^(2(0))]`
= `1/2 ["e"^2 - "e"^0]`
= `1/2 ["e"^2 - 1]`
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^{\pi/2} \frac{1}{5 \cos x + 3 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d\theta\]
\[\int\limits_0^1 \tan^{- 1} x\ dx\]
\[\int\limits_0^{\pi/2} \left( 2 \log \cos x - \log \sin 2x \right) dx\]
\[\int\limits_1^2 \left( x^2 - 1 \right) dx\]
\[\int\limits_a^b x\ dx\]
\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
