What is the integral from 0 to ln(A) of 6e^{2x} ?

The integral from 0 to ln(A) of 6e^{2x} is 3(A^2-1)

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integral from 0 to ln(A) of 6e^{2x}

\int_{0}^{l n (A)} 6 e^{2 x} d x

3 (A^{2} - 1)

Solution steps

\int_{0}^{l n (A)} 6 e^{2 x} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{l n (A)} e^{2 x} d x

Apply u-substitution

: \int_{0}^{2 l n (A)} e^{u} \frac{1}{2} d u

= 6 \cdot \int_{0}^{2 l n (A)} 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

= 6 \cdot \frac{1}{2} \cdot \int_{0}^{2 l n (A)} e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{2} [e^{u}]_{0}^{2 l n (A)}

Simplify

6 \cdot \frac{1}{2} [e^{u}]_{0}^{2 l n (A)} : 3 [e^{u}]_{0}^{2 l n (A)}

= 3 [e^{u}]_{0}^{2 l n (A)}

Compute the boundaries:

A^{2} - 1

= 3 (A^{2} - 1)

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What is the integral from 0 to ln(A) of 6e^{2x} ?
The integral from 0 to ln(A) of 6e^{2x} is 3(A^2-1)