What is the integral from-2 to k of 1/2 e^{2x+4} ?

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

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integral from-2 to k of 1/2 e^{2x+4}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{0}^{2 k + 4} e^{u} \frac{1}{2} d u

= \frac{1}{2} \cdot \int_{0}^{2 k + 4} e^{u} \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{2 k + 4} e^{u} d u

Use the common integral:

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

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

Simplify

\frac{1}{2} \cdot \frac{1}{2} [e^{u}]_{0}^{2 k + 4} : \frac{1}{4} [e^{u}]_{0}^{2 k + 4}

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

Compute the boundaries:

e^{2 k + 4} - 1

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

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