Solve ∫ (from 0 to 4) of (5.75x+1/2x^2)(-1/8x+1) wrt x

$$\int_{ 0 }^{ 4 } (5.75x+ \frac{ 1 }{ 2 } { x }^{ 2 } )(- \frac{ 1 }{ 8 } x+1) d x$$

Evaluate

$$\frac{112}{3}\approx 37.333333333$$

Solution Steps

Use the distributive property to multiply 5.75x+\frac{1}{2}x^{2} by -\frac{1}{8}x+1 and combine like terms.

$$\int _{0}^{4}-\frac{7}{32}x^{2}+5.75x-\frac{1}{16}x^{3}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -\frac{7x^{2}}{32}+\frac{23x}{4}-\frac{x^{3}}{16}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -\frac{7x^{2}}{32}\mathrm{d}x+\int \frac{23x}{4}\mathrm{d}x+\int -\frac{x^{3}}{16}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\frac{7\int x^{2}\mathrm{d}x}{32}+\frac{23\int x\mathrm{d}x}{4}-\frac{\int x^{3}\mathrm{d}x}{16}$$

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 -\frac{7}{32} times \frac{x^{3}}{3}.

$$-\frac{7x^{3}}{96}+\frac{23\int x\mathrm{d}x}{4}-\frac{\int x^{3}\mathrm{d}x}{16}$$

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 5.75 times \frac{x^{2}}{2}.

$$-\frac{7x^{3}}{96}+\frac{23x^{2}}{8}-\frac{\int x^{3}\mathrm{d}x}{16}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -\frac{1}{16} times \frac{x^{4}}{4}.

$$-\frac{7x^{3}}{96}+\frac{23x^{2}}{8}-\frac{x^{4}}{64}$$

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.

$$-\frac{7}{96}\times 4^{3}+\frac{23}{8}\times 4^{2}-\frac{4^{4}}{64}-\left(-\frac{7}{96}\times 0^{3}+\frac{23}{8}\times 0^{2}-\frac{0^{4}}{64}\right)$$

Simplify.

$$\frac{112}{3}$$