What is the integral from 0 to 2 of 5e^x+9cos(x) ?

The integral from 0 to 2 of 5e^x+9cos(x) is 5(e^2-1)+9sin(2)

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integral from 0 to 2 of 5e^x+9cos(x)

\int_{0}^{2} 5 e^{x} + 9 cos (x) d x

5 (e^{2} - 1) + 9 sin (2)

Decimal

40.12895 \dots

Solution steps

\int_{0}^{2} 5 e^{x} + 9 cos (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} 5 e^{x} d x + \int_{0}^{2} 9 cos (x) d x

\int_{0}^{2} 5 e^{x} d x = 5 (e^{2} - 1)

\int_{0}^{2} 9 cos (x) d x = 9 sin (2)

= 5 (e^{2} - 1) + 9 sin (2)

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What is the integral from 0 to 2 of 5e^x+9cos(x) ?
The integral from 0 to 2 of 5e^x+9cos(x) is 5(e^2-1)+9sin(2)