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

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

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integral from 0 to 1 of (4x+2)e^{2x}

\int_{0}^{1} (4 x + 2) e^{2 x} d x

2 e^{2}

Decimal

14.77811 \dots

Solution steps

\int_{0}^{1} (4 x + 2) e^{2 x} d x

(4 x + 2) e^{2 x} = 2 (2 x + 1) e^{2 x}

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

Take the constant out:

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

= 2 \cdot \int_{0}^{1} (2 x + 1) e^{2 x} d x

Apply Integration By Parts

= 2 [e^{2 x} x + \frac{1}{2} e^{2 x} - \int e^{2 x} d x]_{0}^{1}

\int e^{2 x} d x = \frac{1}{2} e^{2 x}

= 2 [e^{2 x} x + \frac{1}{2} e^{2 x} - \frac{1}{2} e^{2 x}]_{0}^{1}

Add similar elements:

\frac{1}{2} e^{2 x} - \frac{1}{2} e^{2 x} = 0

= 2 [e^{2 x} x]_{0}^{1}

Compute the boundaries:

e^{2}

= 2 e^{2}

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