Solve ∫ (from - 0.15 to 665) of [(-x^2+2x)-(-1+1/2x)] wrt x

$$\int _ { - 0.15 } ^ { 665 } [ ( - x ^ { 2 } + 2 x ) - ( - 1 + \frac { 1 } { 2 } x ) ] d x$$

Evaluate

$$-\frac{146541311677}{1500}\approx -97694207.784666667$$

Solution Steps

To find the opposite of -1+\frac{1}{2}x, find the opposite of each term.

$$\int _{-0.15}^{665}-x^{2}+2x+1-\frac{1}{2}x\mathrm{d}x$$

Combine 2x and -\frac{1}{2}x to get \frac{3}{2}x.

$$\int _{-0.15}^{665}-x^{2}+\frac{3}{2}x+1\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -x^{2}+\frac{3x}{2}+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int -x^{2}\mathrm{d}x+\int \frac{3x}{2}\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\int x^{2}\mathrm{d}x+\frac{3\int x\mathrm{d}x}{2}+\int 1\mathrm{d}x$$

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}.

$$-\frac{x^{3}}{3}+\frac{3\int x\mathrm{d}x}{2}+\int 1\mathrm{d}x$$

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 \frac{3}{2} times \frac{x^{2}}{2}.

$$-\frac{x^{3}}{3}+\frac{3x^{2}}{4}+\int 1\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$-\frac{665^{3}}{3}+\frac{3}{4}\times 665^{2}+665-\left(-\frac{\left(-0.15\right)^{3}}{3}+\frac{3}{4}\left(-0.15\right)^{2}-0.15\right)$$

Simplify.

$$-\frac{146541311677}{1500}$$