What is the integral from-1 to 3 of 3xe^{2x^2-5} ?

The integral from-1 to 3 of 3xe^{2x^2-5} is (3(e^{16}-1))/(4e^3)

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integral from-1 to 3 of 3xe^{2x^2-5}

\int_{- 1}^{3} 3 x e^{2 x^{2} - 5} d x

\frac{3 ( e ^{16} - 1 )}{4 e ^{3}}

Decimal

331810.00666 \dots

Solution steps

\int_{- 1}^{3} 3 x e^{2 x^{2} - 5} d x

Take the constant out:

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

= 3 \cdot \int_{- 1}^{3} x e^{2 x^{2} - 5} d x

Apply u-substitution

: \int_{- 3}^{13} \frac{e ^{u}}{4} d u

= 3 \cdot \int_{- 3}^{13} \frac{e ^{u}}{4} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int_{- 3}^{13} e^{u} d u

Use the common integral:

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

= 3 \cdot \frac{1}{4} [e^{u}]_{- 3}^{13}

Simplify

3 \cdot \frac{1}{4} [e^{u}]_{- 3}^{13} : \frac{3}{4} [e^{u}]_{- 3}^{13}

= \frac{3}{4} [e^{u}]_{- 3}^{13}

Compute the boundaries:

e^{13} - \frac{1}{e ^{3}}

= \frac{3}{4} (e^{13} - \frac{1}{e ^{3}})

Simplify

= \frac{3 ( e ^{16} - 1 )}{4 e ^{3}}

x \to 1 lim (\frac{3 x + 7 - 2}{x ^{3} - 1})

\int x + x d x

(\frac{6 x ^{2} - 5}{4 - x})^{3}

x \to 0 lim (\frac{x - 5}{x - 1})

y = (5)^{x}

What is the integral from-1 to 3 of 3xe^{2x^2-5} ?
The integral from-1 to 3 of 3xe^{2x^2-5} is (3(e^{16}-1))/(4e^3)