What is the integral from 1 to ln(2) of 4e^{2x} ?

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

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integral from 1 to ln(2) of 4e^{2x}

\int_{1}^{l n (2)} 4 e^{2 x} d x

- 2 (e^{2} - 4)

Decimal

- 6.77811 \dots

Solution steps

\int_{1}^{l n (2)} 4 e^{2 x} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{l n (2)}^{1} 4 e^{2 x} d x

Take the constant out:

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

= - 4 \cdot \int_{l n (2)}^{1} e^{2 x} d x

Apply u-substitution

: \int_{2 l n (2)}^{2} e^{u} \frac{1}{2} d u

= - 4 \cdot \int_{2 l n (2)}^{2} 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

= - 4 \cdot \frac{1}{2} \cdot \int_{2 l n (2)}^{2} e^{u} d u

Use the common integral:

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

= - 4 \cdot \frac{1}{2} [e^{u}]_{2 l n (2)}^{2}

Simplify

- 4 \cdot \frac{1}{2} [e^{u}]_{2 l n (2)}^{2} : - 2 [e^{u}]_{2 l n (2)}^{2}

= - 2 [e^{u}]_{2 l n (2)}^{2}

Compute the boundaries:

e^{2} - 4

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

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