What is the integral from 0 to 2 of (2x-3e^x) ?

The integral from 0 to 2 of (2x-3e^x) is -3e^2+7

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integral from 0 to 2 of (2x-3e^x)

\int_{0}^{2} (2 x - 3 e^{x}) d x

- 3 e^{2} + 7

Decimal Notation

- 15.16716 \dots

Solution steps

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

\int_{0}^{2} 2 x d x = 4

\int_{0}^{2} 3 e^{x} d x = 3 (e^{2} - 1)

= 4 - 3 (e^{2} - 1)

Simplify

4 - 3 (e^{2} - 1) : - 3 e^{2} + 7

= - 3 e^{2} + 7

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

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

\int_{2}^{8} (5 x^{2} + 3 x + 5) d x

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

\int_{x}^{x^{2}} \frac{1}{t ^{3} + 2} d t

What is the integral from 0 to 2 of (2x-3e^x) ?
The integral from 0 to 2 of (2x-3e^x) is -3e^2+7