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

The integral from 1 to 4 of (4x+2)e^{2x} is 2(4e^8-e^2)

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

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

2 (4 e^{8} - e^{2})

Decimal

23832.88578 \dots

Solution steps

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

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

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

Apply Integration By Parts

: [e^{2 x} x + \frac{1}{2} e^{2 x} - \int e^{2 x} d x]_{1}^{4}

= 2 [e^{2 x} x + \frac{1}{2} e^{2 x} - \int e^{2 x} d x]_{1}^{4}

\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}]_{1}^{4}

Add similar elements:

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

= 2 [e^{2 x} x]_{1}^{4}

Compute the boundaries:

4 e^{8} - e^{2}

= 2 (4 e^{8} - e^{2})

x \to 0 lim (+ (\frac{2}{x} - \frac{3}{x ^{2}}))

\frac{d}{d t} (2 t sin (t) - (t^{2} - 2) cos (t))

\frac{d y}{d x} - \frac{y}{x} = x^{2} y^{2}

\frac{d}{d x} (x^{3} + y^{3} - 3 x y)

x \to - 1 + lim (2 x^{2} - 4 x + 2)

What is the integral from 1 to 4 of (4x+2)e^{2x} ?
The integral from 1 to 4 of (4x+2)e^{2x} is 2(4e^8-e^2)