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

The integral from-1 to 1 of (e^x-e^{2x-1}) is e-1/e-(e^4-1)/(2e^3)

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

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

e - \frac{1}{e} - \frac{e ^{4} - 1}{2 e ^{3}}

Decimal Notation

1.01615 \dots

Solution steps

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

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

\int_{- 1}^{1} e^{2 x - 1} d x = \frac{e ^{4} - 1}{2 e ^{3}}

= e - \frac{1}{e} - \frac{e ^{4} - 1}{2 e ^{3}}

\int_{0}^{1} 2 t^{4} e^{t^{5}} d t

\int_{1}^{3} (- x^{2} + 4) d x

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

\int_{1}^{6} x^{3} d x

\int_{2}^{\infty} \frac{e ^{x}}{x} d x

What is the integral from-1 to 1 of (e^x-e^{2x-1}) ?
The integral from-1 to 1 of (e^x-e^{2x-1}) is e-1/e-(e^4-1)/(2e^3)