Solve ∫ (from 1800 to 1900) of (.95x-1300) wrt x

$$\int_{ 1800 }^{ 1900 } (.95x-1300) d x$$

Evaluate

$$45750$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{19x}{20}-1300\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{19x}{20}\mathrm{d}x+\int -1300\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{19\int x\mathrm{d}x}{20}+\int -1300\mathrm{d}x$$

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 0.95 times \frac{x^{2}}{2}.

$$\frac{19x^{2}}{40}+\int -1300\mathrm{d}x$$

Find the integral of -1300 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{19x^{2}}{40}-1300x$$

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.

$$\frac{19}{40}\times 1900^{2}-1300\times 1900-\left(\frac{19}{40}\times 1800^{2}-1300\times 1800\right)$$

Simplify.

$$45750$$