Advertisements
Advertisements
Question
Evaluate each of the following integral:
Advertisements
Solution
\[I = \int_0^1 x e^{x^2} dx\]
\[ = \frac{1}{2} \int_0^1 e^{x^2} 2xdx\]
Put \[x^2 = z\]
When \[x \to 0, z \to 0\]
When \[x \to 1, z \to 1\]
\[\therefore I = \frac{1}{2} \int_0^1 e^z dz\]
\[ = \frac{1}{2} \left.\times {e^z}\right|_0^1 \]
\[ = \frac{1}{2}\left( e - e^0 \right)\]
\[ = \frac{1}{2}\left( e - 1 \right)\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
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 \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
