What is the integral of 20xe^{0.01x^2} ?

The integral of 20xe^{0.01x^2} is 1000e^{0.01x^2}+C

integral of 20xe^{0.01x^2}

\int 20 x e^{0.01 x^{2}} d x

1000 e^{0.01 x^{2}} + C

Solution steps

\int 20 x e^{0.01 x^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 20 \cdot \int x e^{0.01 x^{2}} d x

Apply u-substitution

= 20 \cdot \int \frac{e ^{u}}{0.02} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 20 \cdot \frac{1}{0.02} \cdot \int e^{u} d u

Simplify

= 20 \cdot 50 \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= 20 \cdot 50 e^{u}

Substitute back

u = 0.01 x^{2}

= 20 \cdot 50 e^{0.01 x^{2}}

Simplify

= 1000 e^{0.01 x^{2}}

Add a constant to the solution

= 1000 e^{0.01 x^{2}} + C

t an g e n t 2 cos (x) + 3 sin (x), at x = \frac{π}{2}

d er i v a t i v e \frac{ln ( 2 x )}{2 x}

\frac{\partial}{\partial x} (\frac{sin ( x )}{cos ( x )})

d er i v a t i v e y = ln (sec (5 x) + tan (5 x))

\frac{d}{d x} (10 sec (2 x) tan (2 x))

What is the integral of 20xe^{0.01x^2} ?
The integral of 20xe^{0.01x^2} is 1000e^{0.01x^2}+C