Solve ∫ (from 0 to 4) of (-70+9.75x)(0.5x) wrt x

$$\int_{ 0 }^{ 4 } (-70+9.75x)(0.5x) d x$$

Evaluate

$$-176$$

Solution Steps

Use the distributive property to multiply -70+9.75x by 0.5.

$$\int _{0}^{4}\left(-35+4.875x\right)x\mathrm{d}x$$

Use the distributive property to multiply -35+4.875x by x.

$$\int _{0}^{4}-35x+4.875x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -35x+\frac{39x^{2}}{8}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -35x\mathrm{d}x+\int \frac{39x^{2}}{8}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-35\int x\mathrm{d}x+\frac{39\int x^{2}\mathrm{d}x}{8}$$

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 -35 times \frac{x^{2}}{2}.

$$-\frac{35x^{2}}{2}+\frac{39\int x^{2}\mathrm{d}x}{8}$$

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 4.875 times \frac{x^{3}}{3}.

$$-\frac{35x^{2}}{2}+\frac{13x^{3}}{8}$$

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.

$$-\frac{35}{2}\times 4^{2}+\frac{13}{8}\times 4^{3}-\left(-\frac{35}{2}\times 0^{2}+\frac{13}{8}\times 0^{3}\right)$$

Simplify.

$$-176$$