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

$$\int_{ 0 }^{ 3 } { \left( { x }^{ 2 } -3x \right) }^{ 2 } d x$$

Evaluate

$$\frac{81}{10}=8.1$$

Solution Steps

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

$$\int _{0}^{3}\left(x^{2}\right)^{2}-6x^{2}x+9x^{2}\mathrm{d}x$$

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

$$\int _{0}^{3}x^{4}-6x^{2}x+9x^{2}\mathrm{d}x$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\int _{0}^{3}x^{4}-6x^{3}+9x^{2}\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -6 times \frac{x^{4}}{4}.

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

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

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

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.

$$3\times 3^{3}-\frac{3}{2}\times 3^{4}+\frac{3^{5}}{5}-\left(3\times 0^{3}-\frac{3}{2}\times 0^{4}+\frac{0^{5}}{5}\right)$$

Simplify.

$$\frac{81}{10}$$