Solve ∫ (from - 2 to 4) of (9s^2+4)ds

$$\int _ { - 2 } ^ { 4 } ( 9 s ^ { 2 } + 4 ) d s$$

Evaluate

$$240$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 9s^{2}+4\mathrm{d}s$$

Integrate the sum term by term.

$$\int 9s^{2}\mathrm{d}s+\int 4\mathrm{d}s$$

Factor out the constant in each of the terms.

$$9\int s^{2}\mathrm{d}s+\int 4\mathrm{d}s$$

Since \int s^{k}\mathrm{d}s=\frac{s^{k+1}}{k+1} for k\neq -1, replace \int s^{2}\mathrm{d}s with \frac{s^{3}}{3}. Multiply 9 times \frac{s^{3}}{3}.

$$3s^{3}+\int 4\mathrm{d}s$$

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

$$3s^{3}+4s$$

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.

$$3\times 4^{3}+4\times 4-\left(3\left(-2\right)^{3}+4\left(-2\right)\right)$$

Simplify.

$$240$$