Solve ∫ (from 0 to 4) of (2.75x+0.5x^2)(-0.5) wrt x

$$\int_{ 0 }^{ 4 } (2.75x+0.5 { x }^{ 2 } )(-0.5) d x$$

Evaluate

$$-\frac{49}{3}\approx -16.333333333$$

Solution Steps

Use the distributive property to multiply 2.75x+0.5x^{2} by -0.5.

$$\int _{0}^{4}-1.375x-0.25x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -\frac{11x}{8}-\frac{x^{2}}{4}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -\frac{11x}{8}\mathrm{d}x+\int -\frac{x^{2}}{4}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\frac{11\int x\mathrm{d}x}{8}-\frac{\int x^{2}\mathrm{d}x}{4}$$

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 -1.375 times \frac{x^{2}}{2}.

$$-\frac{11x^{2}}{16}-\frac{\int x^{2}\mathrm{d}x}{4}$$

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 -0.25 times \frac{x^{3}}{3}.

$$-\frac{11x^{2}}{16}-\frac{x^{3}}{12}$$

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.

$$-\frac{11}{16}\times 4^{2}-\frac{4^{3}}{12}-\left(-\frac{11}{16}\times 0^{2}-\frac{0^{3}}{12}\right)$$

Simplify.

$$-\frac{49}{3}$$