Solve ∫ (from -1 to 3) of 3(-4)left|-4-4right| wrt x=

$$\int_{ -1 }^{ 3 } 3(-4) \left| -4-4 \right| d x =$$

Evaluate

$$-384$$

Solution Steps

Multiply 3 and -4 to get -12.

$$\int _{-1}^{3}-12|-4-4|\mathrm{d}x$$

Subtract 4 from -4 to get -8.

$$\int _{-1}^{3}-12|-8|\mathrm{d}x$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -8 is 8.

$$\int _{-1}^{3}-12\times 8\mathrm{d}x$$

Multiply -12 and 8 to get -96.

$$\int _{-1}^{3}-96\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -96\mathrm{d}x$$

Find the integral of -96 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-96x$$

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.

$$-96\times 3+96\left(-1\right)$$

Simplify.

$$-384$$