What is the integral from 1 to 2 of 2xe^{4x^2+2} ?

The integral from 1 to 2 of 2xe^{4x^2+2} is (e^{18}-e^6)/4

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integral from 1 to 2 of 2xe^{4x^2+2}

\int_{1}^{2} 2 x e^{4 x^{2} + 2} d x

\frac{e ^{18} - e ^{6}}{4}

Decimal

16414891.42713 \dots

Solution steps

\int_{1}^{2} 2 x e^{4 x^{2} + 2} d x

Take the constant out:

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

= 2 \cdot \int_{1}^{2} x e^{4 x^{2} + 2} d x

Apply u-substitution

: \int_{1}^{4} \frac{e ^{4 u + 2}}{2} d u

= 2 \cdot \int_{1}^{4} \frac{e ^{4 u + 2}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{1}^{4} e^{4 u + 2} d u

Apply u-substitution

: \int_{6}^{18} e^{v} \frac{1}{4} d v

= 2 \cdot \frac{1}{2} \cdot \int_{6}^{18} e^{v} \frac{1}{4} d v

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \int_{6}^{18} e^{v} d v

Use the common integral:

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

= 2 \cdot \frac{1}{2} \cdot \frac{1}{4} [e^{v}]_{6}^{18}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{1}{4} [e^{v}]_{6}^{18} : \frac{1}{4} [e^{v}]_{6}^{18}

= \frac{1}{4} [e^{v}]_{6}^{18}

Compute the boundaries:

e^{18} - e^{6}

= \frac{1}{4} (e^{18} - e^{6})

Simplify

= \frac{e ^{18} - e ^{6}}{4}

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What is the integral from 1 to 2 of 2xe^{4x^2+2} ?
The integral from 1 to 2 of 2xe^{4x^2+2} is (e^{18}-e^6)/4