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

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

Evaluate

$$-\frac{7}{10}=-0.7$$

Solution Steps

Subtract 3 from 2 to get -1.

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

Use the distributive property to multiply x by -1-x^{3}.

$$\int _{0}^{1}-x-x^{4}\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -x-x^{4}\mathrm{d}x$$

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

$$-\int x\mathrm{d}x-\int x^{4}\mathrm{d}x$$

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 -1 times \frac{x^{2}}{2}.

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

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 -1 times \frac{x^{5}}{5}.

$$-\frac{x^{2}}{2}-\frac{x^{5}}{5}$$

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.

$$-\frac{1^{2}}{2}-\frac{1^{5}}{5}-\left(-\frac{0^{2}}{2}-\frac{0^{5}}{5}\right)$$

Simplify.

$$-\frac{7}{10}$$