Evaluate
\frac{245}{12}\approx 20.416666667
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\int _{-2}^{5}\left(x^{2}-4x+4\right)\left(3x-1\right)\mathrm{d}x
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.
\int _{-2}^{5}3x^{3}-13x^{2}+16x-4\mathrm{d}x
Use the distributive property to multiply x^{2}-4x+4 by 3x-1 and combine like terms.
\int 3x^{3}-13x^{2}+16x-4\mathrm{d}x
Evaluate the indefinite integral first.
\int 3x^{3}\mathrm{d}x+\int -13x^{2}\mathrm{d}x+\int 16x\mathrm{d}x+\int -4\mathrm{d}x
Integrate the sum term by term.
3\int x^{3}\mathrm{d}x-13\int x^{2}\mathrm{d}x+16\int x\mathrm{d}x+\int -4\mathrm{d}x
Factor out the constant in each of the terms.
\frac{3x^{4}}{4}-13\int x^{2}\mathrm{d}x+16\int x\mathrm{d}x+\int -4\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply 3 times \frac{x^{4}}{4}.
\frac{3x^{4}}{4}-\frac{13x^{3}}{3}+16\int x\mathrm{d}x+\int -4\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 -13 times \frac{x^{3}}{3}.
\frac{3x^{4}}{4}-\frac{13x^{3}}{3}+8x^{2}+\int -4\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 16 times \frac{x^{2}}{2}.
\frac{3x^{4}}{4}-\frac{13x^{3}}{3}+8x^{2}-4x
Find the integral of -4 using the table of common integrals rule \int a\mathrm{d}x=ax.
8x^{2}-4x-\frac{13x^{3}}{3}+\frac{3x^{4}}{4}
Simplify.
8\times 5^{2}-4\times 5-\frac{13}{3}\times 5^{3}+\frac{3}{4}\times 5^{4}-\left(8\left(-2\right)^{2}-4\left(-2\right)-\frac{13}{3}\left(-2\right)^{3}+\frac{3}{4}\left(-2\right)^{4}\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{245}{12}
Simplify.
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