What is the integral from 0 to c of 4e^{2x} ?

The integral from 0 to c of 4e^{2x} is 2(e^{2c}-1)

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integral from 0 to c of 4e^{2x}

\int_{0}^{c} 4 e^{2 x} d x

2 (e^{2 c} - 1)

Solution steps

\int_{0}^{c} 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_{0}^{c} e^{2 x} d x

Apply u-substitution

: \int_{0}^{2 c} e^{u} \frac{1}{2} d u

= 4 \cdot \int_{0}^{2 c} 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_{0}^{2 c} e^{u} d u

Use the common integral:

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

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

Simplify

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

= 2 [e^{u}]_{0}^{2 c}

Compute the boundaries:

e^{2 c} - 1

= 2 (e^{2 c} - 1)

\int_{0}^{1} (e^{x} - x e^{x^{2}}) d x

\int \frac{v}{1 - v} d v

\int - x + 3 (cos (\frac{n \cdot π \cdot x}{8})) d x

\frac{d}{d x} (\frac{1}{x y})

(x^{2} - x y) \frac{d y}{d x} + y^{2} = 0

What is the integral from 0 to c of 4e^{2x} ?
The integral from 0 to c of 4e^{2x} is 2(e^{2c}-1)