Solve ∫ (from 0 to 4) of (-72x+11x)(0.5x) wrt x

$$\int_{ 0 }^{ 4 } (-72x+11x)(0.5x) d x$$

Evaluate

$$-\frac{1952}{3}\approx -650.666666667$$

Solution Steps

Combine -72x and 11x to get -61x.

$$\int _{0}^{4}-61x\times 0.5x\mathrm{d}x$$

Multiply -61 and 0.5 to get -30.5.

$$\int _{0}^{4}-30.5xx\mathrm{d}x$$

Multiply x and x to get x^{2}.

$$\int _{0}^{4}-30.5x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -\frac{61x^{2}}{2}\mathrm{d}x$$

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

$$-\frac{61\int x^{2}\mathrm{d}x}{2}$$

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}.

$$-\frac{61x^{3}}{6}$$

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.

$$-\frac{61}{6}\times 4^{3}+\frac{61}{6}\times 0^{3}$$

Simplify.

$$-\frac{1952}{3}$$