Solve ∫ (from 0 to 4) of 32(x-2) wrt x

$$\int _ { 0 } ^ { 4 } 32 ( x - 2 ) d x$$

Evaluate

$$0$$

Solution Steps

Use the distributive property to multiply 32 by x-2.

$$\int _{0}^{4}32x-64\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 32x-64\mathrm{d}x$$

Integrate the sum term by term.

$$\int 32x\mathrm{d}x+\int -64\mathrm{d}x$$

Factor out the constant in each of the terms.

$$32\int x\mathrm{d}x+\int -64\mathrm{d}x$$

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

$$16x^{2}+\int -64\mathrm{d}x$$

Find the integral of -64 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$16x^{2}-64x$$

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.

$$16\times 4^{2}-64\times 4-\left(16\times 0^{2}-64\times 0\right)$$

Simplify.

$$0$$