What is the integral from 0 to 1 of te^{2t} ?

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

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integral from 0 to 1 of te^{2t}

\int_{0}^{1} t e^{2 t} d t

\frac{e ^{2} + 1}{4}

Decimal Notation

2.09726 \dots

Solution steps

\int_{0}^{1} t e^{2 t} d t

Apply u-substitution

: \int_{0}^{2} \frac{e ^{u} u}{4} d u

= \int_{0}^{2} \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}^{2} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{2}

= \frac{1}{4} [e^{u} u - \int e^{u} d u]_{0}^{2}

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

= \frac{1}{4} [e^{u} u - e^{u}]_{0}^{2}

Compute the boundaries:

e^{2} + 1

= \frac{1}{4} (e^{2} + 1)

Simplify

= \frac{e ^{2} + 1}{4}

tan (x), at x = - \frac{π}{3}

\int_{0}^{14} 10 e^{x} d x

\frac{\partial}{\partial x} (x y ln (\frac{x}{y}))

x \to - \frac{3 π}{2} - lim (tan (x))

\int_{1}^{b} \frac{1}{( x + 7 ) ^{2}} d x

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