What is the integral from-1 to 3 of (e^x-1) ?

The integral from-1 to 3 of (e^x-1) is e^3-1/e-4

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integral from-1 to 3 of (e^x-1)

\int_{- 1}^{3} (e^{x} - 1) d x

e^{3} - \frac{1}{e} - 4

Decimal

15.71765 \dots

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{3} e^{x} d x - \int_{- 1}^{3} 1 d x

\int_{- 1}^{3} e^{x} d x = e^{3} - \frac{1}{e}

\int_{- 1}^{3} 1 d x = 4

= e^{3} - \frac{1}{e} - 4

\int_{1}^{e^{4}} \frac{4 ln ( x )}{x} d x

\int_{- 2}^{1} (3 - y) - (y^{2} + 1) d y

\int_{0}^{\infty} 9 x e^{- 3 x} d x

\int_{\frac{1}{2}}^{1} x d x

\int_{2}^{\infty} \frac{1}{x ^{2} + 1} d x

What is the integral from-1 to 3 of (e^x-1) ?
The integral from-1 to 3 of (e^x-1) is e^3-1/e-4