Solve ∫ (from - 3 to 4) of (x-1)^2(4x+2) wrt x=

$$\int _ { - 3 } ^ { 4 } ( x - 1 ) ^ { 2 } ( 4 x + 2 ) d x =$$

Evaluate

$$7$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$\int _{-3}^{4}\left(x^{2}-2x+1\right)\left(4x+2\right)\mathrm{d}x$$

Use the distributive property to multiply x^{2}-2x+1 by 4x+2 and combine like terms.

$$\int _{-3}^{4}4x^{3}-6x^{2}+2\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 4x^{3}-6x^{2}+2\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{3}\mathrm{d}x+\int -6x^{2}\mathrm{d}x+\int 2\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{3}\mathrm{d}x-6\int x^{2}\mathrm{d}x+\int 2\mathrm{d}x$$

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

$$x^{4}-6\int x^{2}\mathrm{d}x+\int 2\mathrm{d}x$$

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 -6 times \frac{x^{3}}{3}.

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

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

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

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.

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

Simplify.

$$7$$