What is the integral from 0 to 1 of ye^{2y} ?

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

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integral from 0 to 1 of ye^{2y}

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

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

Decimal

2.09726 \dots

Solution steps

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

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}

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