Solve ∫ (from 0 to 4) of (-5(2+x))(-1x) wrt x

$$\int_{ 0 }^{ 4 } (-5(2+x))(-1x) d x$$

Evaluate

$$\frac{560}{3}\approx 186.666666667$$

Solution Steps

Multiply -5 and -1 to get 5.

$$\int _{0}^{4}5\left(2+x\right)x\mathrm{d}x$$

Use the distributive property to multiply 5 by 2+x.

$$\int _{0}^{4}\left(10+5x\right)x\mathrm{d}x$$

Use the distributive property to multiply 10+5x by x.

$$\int _{0}^{4}10x+5x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 10x+5x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 10x\mathrm{d}x+\int 5x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$10\int x\mathrm{d}x+5\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 10 times \frac{x^{2}}{2}.

$$5x^{2}+5\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 5 times \frac{x^{3}}{3}.

$$5x^{2}+\frac{5x^{3}}{3}$$

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.

$$5\times 4^{2}+\frac{5}{3}\times 4^{3}-\left(5\times 0^{2}+\frac{5}{3}\times 0^{3}\right)$$

Simplify.

$$\frac{560}{3}$$