What is the integral from 1 to 2 of 16xe^{4x^2} ?

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

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integral from 1 to 2 of 16xe^{4x^2}

\int_{1}^{2} 16 x e^{4 x^{2}} d x

2 (e^{16} - e^{4})

Decimal Notation

17772111.84471 \dots

Solution steps

\int_{1}^{2} 16 x e^{4 x^{2}} d x

Take the constant out:

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

= 16 \cdot \int_{1}^{2} x e^{4 x^{2}} d x

Apply u-substitution

: \int_{4}^{16} \frac{e ^{u}}{8} d u

= 16 \cdot \int_{4}^{16} \frac{e ^{u}}{8} d u

Take the constant out:

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

= 16 \cdot \frac{1}{8} \cdot \int_{4}^{16} e^{u} d u

Use the common integral:

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

= 16 \cdot \frac{1}{8} [e^{u}]_{4}^{16}

Simplify

16 \cdot \frac{1}{8} [e^{u}]_{4}^{16} : 2 [e^{u}]_{4}^{16}

= 2 [e^{u}]_{4}^{16}

Compute the boundaries:

e^{16} - e^{4}

= 2 (e^{16} - e^{4})

\int_{0}^{1} \frac{( e ^{x} + 1 )}{e ^{x}} d x

\int_{- 1}^{2} \frac{1}{x ^{2} - 1} d x

\int_{0}^{2} x^{3} e^{- x^{2}} d x

\int_{0}^{2} \frac{y}{2} d y

\int_{1}^{4} (x^{2} - 1) d x

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