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

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

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

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

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

Decimal Notation

- 3.62686 \dots

Solution steps

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

\int_{- 1}^{1} 4 x^{3} d x = 0

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

= 0 - \frac{e ^{4} - 1}{2 e ^{2}}

Simplify

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

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