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

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

Evaluate

$$-\frac{7}{4}=-1.75$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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 -1 times \frac{x^{4}}{4}.

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

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.

$$2\times 2-\frac{2^{4}}{4}-\left(2\times 1-\frac{1^{4}}{4}\right)$$

Simplify.

$$-\frac{7}{4}$$