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

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

Evaluate

$$\frac{416}{35}\approx 11.885714286$$

Solution Steps

Square x^{3}-3x+2.

$$\int _{-1}^{1}x^{6}-6x^{4}+4x^{3}+9x^{2}-12x+4\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{6}-6x^{4}+4x^{3}+9x^{2}-12x+4\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{6}\mathrm{d}x+\int -6x^{4}\mathrm{d}x+\int 4x^{3}\mathrm{d}x+\int 9x^{2}\mathrm{d}x+\int -12x\mathrm{d}x+\int 4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{6}\mathrm{d}x-6\int x^{4}\mathrm{d}x+4\int x^{3}\mathrm{d}x+9\int x^{2}\mathrm{d}x-12\int x\mathrm{d}x+\int 4\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}.

$$\frac{x^{7}}{7}-6\int x^{4}\mathrm{d}x+4\int x^{3}\mathrm{d}x+9\int x^{2}\mathrm{d}x-12\int x\mathrm{d}x+\int 4\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply -6 times \frac{x^{5}}{5}.

$$\frac{x^{7}}{7}-\frac{6x^{5}}{5}+4\int x^{3}\mathrm{d}x+9\int x^{2}\mathrm{d}x-12\int x\mathrm{d}x+\int 4\mathrm{d}x$$

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

$$\frac{x^{7}}{7}-\frac{6x^{5}}{5}+x^{4}+9\int x^{2}\mathrm{d}x-12\int x\mathrm{d}x+\int 4\mathrm{d}x$$

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

$$\frac{x^{7}}{7}-\frac{6x^{5}}{5}+x^{4}+3x^{3}-12\int x\mathrm{d}x+\int 4\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -12 times \frac{x^{2}}{2}.

$$\frac{x^{7}}{7}-\frac{6x^{5}}{5}+x^{4}+3x^{3}-6x^{2}+\int 4\mathrm{d}x$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{7}}{7}-\frac{6x^{5}}{5}+x^{4}+3x^{3}-6x^{2}+4x$$

Simplify.

$$-6x^{2}+\frac{x^{7}}{7}-\frac{6x^{5}}{5}+x^{4}+3x^{3}+4x$$

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.

$$-6\times 1^{2}+\frac{1^{7}}{7}-\frac{6}{5}\times 1^{5}+1^{4}+3\times 1^{3}+4\times 1-\left(-6\left(-1\right)^{2}+\frac{\left(-1\right)^{7}}{7}-\frac{6}{5}\left(-1\right)^{5}+\left(-1\right)^{4}+3\left(-1\right)^{3}+4\left(-1\right)\right)$$

Simplify.

$$\frac{416}{35}$$