Solve ∫ (from 0 to 3) of (x^3-4x)^2 wrt x

$$\int _ { 0 } ^ { 3 } ( x ^ { 3 } - 4 x ) ^ { 2 } d x$$

Evaluate

$$\frac{2367}{35}\approx 67.628571429$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{3}-4x\right)^{2}.

$$\int _{0}^{3}\left(x^{3}\right)^{2}-8x^{3}x+16x^{2}\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$\int _{0}^{3}x^{6}-8x^{3}x+16x^{2}\mathrm{d}x$$

To multiply powers of the same base, add their exponents. Add 3 and 1 to get 4.

$$\int _{0}^{3}x^{6}-8x^{4}+16x^{2}\mathrm{d}x$$

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.

$$\frac{16}{3}\times 3^{3}-\frac{8}{5}\times 3^{5}+\frac{3^{7}}{7}-\left(\frac{16}{3}\times 0^{3}-\frac{8}{5}\times 0^{5}+\frac{0^{7}}{7}\right)$$

Simplify.

$$\frac{2367}{35}$$