Solve ∫ (from 0 to 50) of [30-0.05x^2] wrt x

$$\int _ { 0 } ^ { 50 } [ 30 - 0.05 x ^ { 2 } ] d x$$

Evaluate

$$-\frac{1750}{3}\approx -583.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 30-\frac{x^{2}}{20}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 30\mathrm{d}x+\int -\frac{x^{2}}{20}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 30\mathrm{d}x-\frac{\int x^{2}\mathrm{d}x}{20}$$

Find the integral of 30 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$30x-\frac{\int x^{2}\mathrm{d}x}{20}$$

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 -0.05 times \frac{x^{3}}{3}.

$$30x-\frac{x^{3}}{60}$$

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.

$$30\times 50-\frac{50^{3}}{60}-\left(30\times 0-\frac{0^{3}}{60}\right)$$

Simplify.

$$-\frac{1750}{3}$$