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

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

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integral from 0 to s of 2e^x

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

2 (e^{s} - 1)

Solution steps

\int_{0}^{s} 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}^{s} e^{x} d x

Use the common integral:

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

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

Compute the boundaries:

e^{s} - 1

= 2 (e^{s} - 1)

\int_{0}^{1} 5 x + 3 d x

\int_{0}^{2} 2 x^{3} + 3 d x

\int_{- \frac{π}{2}}^{π} cos (x) d x

\int_{- 2}^{0} - cos (\frac{nπ x}{2}) d x

\int_{1}^{1} x^{3} d x

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