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

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

Evaluate

$$\frac{1625}{21}\approx 77.380952381$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-4\right)^{2}.

$$\int _{-3}^{2}x^{2}\left(\left(x^{2}\right)^{2}-8x^{2}+16\right)\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\int _{-3}^{2}x^{2}\left(x^{4}-8x^{2}+16\right)\mathrm{d}x$$

Use the distributive property to multiply x^{2} by x^{4}-8x^{2}+16.

$$\int _{-3}^{2}x^{6}-8x^{4}+16x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{6}-8x^{4}+16x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{6}\mathrm{d}x+\int -8x^{4}\mathrm{d}x+\int 16x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{6}\mathrm{d}x-8\int x^{4}\mathrm{d}x+16\int x^{2}\mathrm{d}x$$

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}.

$$\frac{x^{7}}{7}-8\int x^{4}\mathrm{d}x+16\int x^{2}\mathrm{d}x$$

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 -8 times \frac{x^{5}}{5}.

$$\frac{x^{7}}{7}-\frac{8x^{5}}{5}+16\int x^{2}\mathrm{d}x$$

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

$$\frac{x^{7}}{7}-\frac{8x^{5}}{5}+\frac{16x^{3}}{3}$$

Simplify.

$$\frac{16x^{3}}{3}-\frac{8x^{5}}{5}+\frac{x^{7}}{7}$$

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.

$$\frac{16}{3}\times 2^{3}-\frac{8}{5}\times 2^{5}+\frac{2^{7}}{7}-\left(\frac{16}{3}\left(-3\right)^{3}-\frac{8}{5}\left(-3\right)^{5}+\frac{\left(-3\right)^{7}}{7}\right)$$

Simplify.

$$\frac{1625}{21}$$