Solve ∫ (from 0 to 3) of (x^4-6x^3+9x^2) wrt x

$$\int_{ 0 }^{ 3 } ( { x }^{ 4 } -6 { x }^{ 3 } +9 { x }^{ 2 } ) d x$$

Evaluate

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

Solution Steps

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.

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

Simplify.

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