Solve ∫ (from 3 to 6) of (x-4x) wrt x

$$\int_{ 3 }^{ 6 } (x-4x) d x$$

Evaluate

$$-\frac{81}{2}=-40.5$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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 -4 times \frac{x^{2}}{2}.

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

Simplify.

$$-\frac{3x^{2}}{2}$$

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.

$$-\frac{3}{2}\times 6^{2}+\frac{3}{2}\times 3^{2}$$

Simplify.

$$-\frac{81}{2}$$