Solve ∫ (from -1 to 3) of 3(-20)left|-20-4right| wrt x=

$$\int_{ -1 }^{ 3 } 3(-20) \left| -20-4 \right| d x =$$

Evaluate

$$-5760$$

Solution Steps

Multiply 3 and -20 to get -60.

$$\int _{-1}^{3}-60|-20-4|\mathrm{d}x$$

Subtract 4 from -20 to get -24.

$$\int _{-1}^{3}-60|-24|\mathrm{d}x$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -24 is 24.

$$\int _{-1}^{3}-60\times 24\mathrm{d}x$$

Multiply -60 and 24 to get -1440.

$$\int _{-1}^{3}-1440\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -1440\mathrm{d}x$$

Find the integral of -1440 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-1440x$$

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.

$$-1440\times 3+1440\left(-1\right)$$

Simplify.

$$-5760$$