Solve ∫ (from 0 to 4) of (31+9.75x)(-2+0.5x) wrt x

$$\int_{ 0 }^{ 4 } (31+9.75x)(-2+0.5x) d x$$

Evaluate

$$-176$$

Solution Steps

Apply the distributive property by multiplying each term of 31+9.75x by each term of -2+0.5x.

$$\int _{0}^{4}-62+15.5x-19.5x+4.875x^{2}\mathrm{d}x$$

Combine 15.5x and -19.5x to get -4x.

$$\int _{0}^{4}-62-4x+4.875x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -62-4x+\frac{39x^{2}}{8}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -62\mathrm{d}x+\int -4x\mathrm{d}x+\int \frac{39x^{2}}{8}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int -62\mathrm{d}x-4\int x\mathrm{d}x+\frac{39\int x^{2}\mathrm{d}x}{8}$$

Find the integral of -62 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-62x-4\int x\mathrm{d}x+\frac{39\int x^{2}\mathrm{d}x}{8}$$

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 -4 times \frac{x^{2}}{2}.

$$-62x-2x^{2}+\frac{39\int x^{2}\mathrm{d}x}{8}$$

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}.

$$-62x-2x^{2}+\frac{13x^{3}}{8}$$

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.

$$-62\times 4-2\times 4^{2}+\frac{13}{8}\times 4^{3}-\left(-62\times 0-2\times 0^{2}+\frac{13}{8}\times 0^{3}\right)$$

Simplify.

$$-176$$