Solve ∫ (from 1 to 4) of 4-x-2-x^2 wrt x

$$\int_{ 1 }^{ 4 } 4-x-2- { x }^{ 2 } d x$$

Evaluate

$$-\frac{45}{2}=-22.5$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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

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

Find the integral of -2 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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 -1 times \frac{x^{3}}{3}.

$$4x-\frac{x^{2}}{2}-2x-\frac{x^{3}}{3}$$

Simplify.

$$2x-\frac{x^{2}}{2}-\frac{x^{3}}{3}$$

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.

$$2\times 4-\frac{4^{2}}{2}-\frac{4^{3}}{3}-\left(2\times 1-\frac{1^{2}}{2}-\frac{1^{3}}{3}\right)$$

Simplify.

$$-\frac{45}{2}$$