Solve ∫ (from 0 to a) of 10-0.02x wrt x

$$\int _ { 0 } ^ { a } 10 - 0.02 x d x$$

Evaluate

$$-\frac{a^{2}}{100}+10a$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 10-\frac{x}{50}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 10\mathrm{d}x+\int -\frac{x}{50}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 10\mathrm{d}x-\frac{\int x\mathrm{d}x}{50}$$

Find the integral of 10 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$10x-\frac{\int x\mathrm{d}x}{50}$$

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 -0.02 times \frac{x^{2}}{2}.

$$10x-\frac{x^{2}}{100}$$

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.

$$10a-\frac{a^{2}}{100}-\left(10\times 0-\frac{0^{2}}{100}\right)$$

Simplify.

$$10a-\frac{a^{2}}{100}$$

Differentiate w.r.t. a

$$-\frac{a}{50}+10$$