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\int _{-0.15}^{6.65}-x^{2}+2x+1-\frac{1}{2}x\mathrm{d}x
To find the opposite of -1+\frac{1}{2}x, find the opposite of each term.
\int _{-0.15}^{6.65}-x^{2}+\frac{3}{2}x+1\mathrm{d}x
Combine 2x and -\frac{1}{2}x to get \frac{3}{2}x.
\int -x^{2}+\frac{3x}{2}+1\mathrm{d}x
Evaluate the indefinite integral first.
\int -x^{2}\mathrm{d}x+\int \frac{3x}{2}\mathrm{d}x+\int 1\mathrm{d}x
Integrate the sum term by term.
-\int x^{2}\mathrm{d}x+\frac{3\int x\mathrm{d}x}{2}+\int 1\mathrm{d}x
Factor out the constant in each of the terms.
-\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^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -1 times \frac{x^{3}}{3}.
-\frac{x^{3}}{3}+\frac{3x^{2}}{4}+\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}+x
Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.
-\frac{6.65^{3}}{3}+\frac{3}{4}\times 6.65^{2}+6.65-\left(-\frac{\left(-0.15\right)^{3}}{3}+\frac{3}{4}\left(-0.15\right)^{2}-0.15\right)
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{174233}{3000}
Simplify.