Solve ∫ (from 1 to 2) of 4-x^2 wrt x

$$\int_{ 1 }^{ 2 } 4- { x }^{ 2 } d x$$

Evaluate

$$\frac{5}{3}\approx 1.666666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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 -1 times \frac{x^{3}}{3}.

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

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.

$$4\times 2-\frac{2^{3}}{3}-\left(4\times 1-\frac{1^{3}}{3}\right)$$

Simplify.

$$\frac{5}{3}$$