Solve ∫ (from 0 to 5) of (2e^x+4cosx) wrt x

$$\int _ { 0 } ^ { 5 } ( 2 e ^ { x } + 4 \cos x ) d x$$

Evaluate

$$2\left(e^{5}+2\sin(5)-1\right)\approx 290.990621107$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 2\left(e^{x}+2\cos(x)\right)\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2e^{x}\mathrm{d}x+\int 4\cos(x)\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\left(\int e^{x}\mathrm{d}x+2\int \cos(x)\mathrm{d}x\right)$$

Use \int e^{x}\mathrm{d}x=e^{x} from the table of common integrals to obtain the result.

$$2\left(e^{x}+2\int \cos(x)\mathrm{d}x\right)$$

Use \int \cos(x)\mathrm{d}x=\sin(x) from the table of common integrals to obtain the result.

$$2\left(e^{x}+2\sin(x)\right)$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$2e^{5}+4\sin(5)-\left(2e^{0}+4\sin(0)\right)$$

Simplify.

$$2\left(e^{5}+2\sin(5)-1\right)$$