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

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

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

\int_{0}^{1} e^{2 x} d x

e - 1

Decimal

1.71828 \dots

Solution steps

\int_{0}^{1} e^{2 x} d x

Simplify

e^{2 x} : e^{x}

= \int_{0}^{1} e^{x} d x

Use the common integral:

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

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

Compute the boundaries:

e - 1

= e - 1

15 - x^{2}, x^{2} - 3

n = 1 \sum \infty 2 e^{- 9 n}

\int t e^{8 t} d t

f (x) = 5 x^{- \frac{3}{5}}

y = x^{3} - 7 x^{2} + 8 x + 1

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