Solve ∫ (from -1 to 7) of 2x+5-(x^2-2x) wrt x

$$\int_{ -1 }^{ 7 } 2x+5-( { x }^{ 2 } -2x) d x$$

Evaluate

$$\frac{64}{3}\approx 21.333333333$$

Solution Steps

To find the opposite of x^{2}-2x, find the opposite of each term.

$$\int _{-1}^{7}2x+5-x^{2}+2x\mathrm{d}x$$

Combine 2x and 2x to get 4x.

$$\int _{-1}^{7}4x+5-x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 4x+5-x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x\mathrm{d}x+\int 5\mathrm{d}x+\int -x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x\mathrm{d}x+\int 5\mathrm{d}x-\int x^{2}\mathrm{d}x$$

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

$$2x^{2}+\int 5\mathrm{d}x-\int x^{2}\mathrm{d}x$$

Find the integral of 5 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{2}+5x-\int x^{2}\mathrm{d}x$$

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}.

$$2x^{2}+5x-\frac{x^{3}}{3}$$

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.

$$2\times 7^{2}+5\times 7-\frac{7^{3}}{3}-\left(2\left(-1\right)^{2}+5\left(-1\right)-\frac{\left(-1\right)^{3}}{3}\right)$$

Simplify.

$$\frac{64}{3}$$