What is the integral from 0 to 2 of 9x^2-e^x ?

The integral from 0 to 2 of 9x^2-e^x is -e^2+25

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integral from 0 to 2 of 9x^2-e^x

\int_{0}^{2} 9 x^{2} - e^{x} d x

- e^{2} + 25

Decimal Notation

17.61094 \dots

Solution steps

\int_{0}^{2} 9 x^{2} - 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} 9 x^{2} d x - \int_{0}^{2} e^{x} d x

\int_{0}^{2} 9 x^{2} d x = 24

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

= 24 - (e^{2} - 1)

Simplify

= - e^{2} + 25

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What is the integral from 0 to 2 of 9x^2-e^x ?
The integral from 0 to 2 of 9x^2-e^x is -e^2+25