Solve ∫ (from 0 to 4) of 5*4-4^2

$$\int _ { 0 } ^ { 4 } 5 \cdot 4 - 4 ^ { 2 }$$

Evaluate

$$16$$

Solution Steps

Multiply 5 and 4 to get 20.

$$\int _{0}^{4}20-4^{2}\mathrm{d}x$$

Calculate 4 to the power of 2 and get 16.

$$\int _{0}^{4}20-16\mathrm{d}x$$

Subtract 16 from 20 to get 4.

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

Evaluate the indefinite integral first.

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

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$4x$$

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.

$$4\times 4-4\times 0$$

Simplify.

$$16$$