Solve ∫ (from 0 to 2) of left(x-1right)^3-(x-1) wrt x

$$\int_{ 0 }^{ 2 } { \left(x-1 \right) }^{ 3 } -(x-1) d x$$

Evaluate

$$0$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-1\right)^{3}.

$$\int _{0}^{2}x^{3}-3x^{2}+3x-1-\left(x-1\right)\mathrm{d}x$$

To find the opposite of x-1, find the opposite of each term.

$$\int _{0}^{2}x^{3}-3x^{2}+3x-1-x+1\mathrm{d}x$$

Combine 3x and -x to get 2x.

$$\int _{0}^{2}x^{3}-3x^{2}+2x-1+1\mathrm{d}x$$

Add -1 and 1 to get 0.

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

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

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

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.

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

Simplify.

$$0$$