What is the integral from 0 to 2 of te^{2t} ?

The integral from 0 to 2 of te^{2t} is (3e^4+1)/4

Popular >

integral from 0 to 2 of te^{2t}

\int_{0}^{2} t e^{2 t} d t

\frac{3 e ^{4} + 1}{4}

Decimal Notation

41.19861 \dots

Solution steps

\int_{0}^{2} t e^{2 t} d t

Apply u-substitution

: \int_{0}^{4} \frac{e ^{u} u}{4} d u

= \int_{0}^{4} \frac{e ^{u} u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{4} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{4}

= \frac{1}{4} [e^{u} u - \int e^{u} d u]_{0}^{4}

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

= \frac{1}{4} [e^{u} u - e^{u}]_{0}^{4}

Compute the boundaries:

3 e^{4} + 1

= \frac{1}{4} (3 e^{4} + 1)

Simplify

= \frac{3 e ^{4} + 1}{4}

\frac{\partial}{\partial x} (x^{2} e^{3 x y})

\int (\frac{x}{x x ^{2} + 36}) d x

\frac{\partial}{\partial y} (ln (x + y^{2}))

\int_{0}^{5} \frac{1}{x ^{\frac{20}{21}}} d x

\int tan (2 x) sec^{8} (2 x) d x

What is the integral from 0 to 2 of te^{2t} ?
The integral from 0 to 2 of te^{2t} is (3e^4+1)/4