What is the integral from 0 to b of 2e^x ?

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

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integral from 0 to b of 2e^x

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

2 (e^{b} - 1)

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

e^{b} - 1

= 2 (e^{b} - 1)

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

\int_{0}^{\frac{π}{2}} sin^{2} (y) d y

\int_{0}^{1} - x^{3} + x d x

\int_{0}^{1} e^{t} e^{e^{t}} d t

\int_{0}^{3} (\frac{2}{x + 1})^{2} d x

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