What is the integral from 0 to 5 of 2e^x+4cos(x) ?

The integral from 0 to 5 of 2e^x+4cos(x) is 2(e^5-1)+4sin(5)

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integral from 0 to 5 of 2e^x+4cos(x)

\int_{0}^{5} 2 e^{x} + 4 cos (x) d x

2 (e^{5} - 1) + 4 sin (5)

Decimal

290.99062 \dots

Solution steps

\int_{0}^{5} 2 e^{x} + 4 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}^{5} 2 e^{x} d x + \int_{0}^{5} 4 cos (x) d x

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

\int_{0}^{5} 4 cos (x) d x = 4 sin (5)

= 2 (e^{5} - 1) + 4 sin (5)

\int_{1}^{2} \frac{9}{x ^{2}} d x

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\int_{0}^{π} x^{2} cos (n) x d x

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

\int_{- \frac{π}{3}}^{\frac{2 π}{3}} 3 tan (x) d x

What is the integral from 0 to 5 of 2e^x+4cos(x) ?
The integral from 0 to 5 of 2e^x+4cos(x) is 2(e^5-1)+4sin(5)