What is the integral from 0 to 1 of xe^{2x^2} ?

The integral from 0 to 1 of xe^{2x^2} is (e^2-1)/4

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integral from 0 to 1 of xe^{2x^2}

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

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

Decimal

1.59726 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{2} - 1

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

Simplify

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

n = 1 \sum \infty p (1 - p)^{2 n}

\frac{d y}{d x} = \frac{y ^{2} + 1}{x ^{2} + 1}, y (0) = 1

\int \frac{3 x ^{7}}{4 + x ^{8}} d x

f (x) = 5 lo g_{10} (4 x^{2})

\frac{d}{d x} (sin^{4} (2 x) cos^{4} (3 x))

What is the integral from 0 to 1 of xe^{2x^2} ?
The integral from 0 to 1 of xe^{2x^2} is (e^2-1)/4