Solve ∫ (from -1 to 1) of left(x^3+x^2+1right)^2-x^6-{x}^4 wrt x

$$\int_{ -1 }^{ 1 } { \left( { x }^{ 3 } + { x }^{ 2 } +1 \right) }^{ 2 } - { x }^{ 6 } - { x }^{ 4 } d x$$

Evaluate

$$\frac{10}{3}\approx 3.333333333$$

Solution Steps

Square x^{3}+x^{2}+1.

$$\int _{-1}^{1}x^{6}+2x^{5}+x^{4}+2x^{3}+2x^{2}+1-x^{6}-x^{4}\mathrm{d}x$$

Combine x^{6} and -x^{6} to get 0.

$$\int _{-1}^{1}2x^{5}+x^{4}+2x^{3}+2x^{2}+1-x^{4}\mathrm{d}x$$

Combine x^{4} and -x^{4} to get 0.

$$\int _{-1}^{1}2x^{5}+2x^{3}+2x^{2}+1\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 2x^{5}+2x^{3}+2x^{2}+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{5}\mathrm{d}x+\int 2x^{3}\mathrm{d}x+\int 2x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{5}\mathrm{d}x+2\int x^{3}\mathrm{d}x+2\int x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. Multiply 2 times \frac{x^{6}}{6}.

$$\frac{x^{6}}{3}+2\int x^{3}\mathrm{d}x+2\int x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply 2 times \frac{x^{4}}{4}.

$$\frac{x^{6}}{3}+\frac{x^{4}}{2}+2\int x^{2}\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$\frac{x^{6}}{3}+\frac{x^{4}}{2}+\frac{2x^{3}}{3}+\int 1\mathrm{d}x$$

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

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

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.

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

Simplify.

$$\frac{10}{3}$$