What is the integral from 0 to 1 of 7xe^{x^2} ?

The integral from 0 to 1 of 7xe^{x^2} is (7(e-1))/2

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integral from 0 to 1 of 7xe^{x^2}

\int_{0}^{1} 7 x e^{x^{2}} d x

\frac{7 ( e - 1 )}{2}

Decimal Notation

6.01398 \dots

Solution steps

\int_{0}^{1} 7 x e^{x^{2}} d x

Take the constant out:

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

= 7 \cdot \int_{0}^{1} x e^{x^{2}} d x

Apply u-substitution

: \int_{0}^{1} \frac{e ^{u}}{2} d u

= 7 \cdot \int_{0}^{1} \frac{e ^{u}}{2} d u

Take the constant out:

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

= 7 \cdot \frac{1}{2} \cdot \int_{0}^{1} e^{u} d u

Use the common integral:

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

= 7 \cdot \frac{1}{2} [e^{u}]_{0}^{1}

Simplify

7 \cdot \frac{1}{2} [e^{u}]_{0}^{1} : \frac{7}{2} [e^{u}]_{0}^{1}

= \frac{7}{2} [e^{u}]_{0}^{1}

Compute the boundaries:

e - 1

= \frac{7}{2} (e - 1)

Simplify

= \frac{7 ( e - 1 )}{2}

y^{^{''}} - 6 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

2 (3 - x^{2})^{7}

\frac{d}{d x} (cos (2 x))

\int \frac{x}{3 - 4 x ^{2}} d x

x \to \infty lim (91)

What is the integral from 0 to 1 of 7xe^{x^2} ?
The integral from 0 to 1 of 7xe^{x^2} is (7(e-1))/2