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

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

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

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

1

Solution steps

\int_{0}^{1} (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]_{0}^{1}

= [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}

= [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} = - e^{2 x}

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

Compute the boundaries:

1

= 1

\int 45 sin (3) (x) cos (2) (x) d x

x^{3}

x \to 1 + lim (e^{x})

x^{\frac{4}{7}}

x \to 3 lim ((x) + 3 x)

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