Solve ∫ (from 0 to 3) of (75t^4+24t^2)dt

$$\int _ { 0 } ^ { 3 } ( 75 t ^ { 4 } + 24 t ^ { 2 } ) d t$$

Evaluate

$$3861$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 75t^{4}+24t^{2}\mathrm{d}t$$

Integrate the sum term by term.

$$\int 75t^{4}\mathrm{d}t+\int 24t^{2}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$75\int t^{4}\mathrm{d}t+24\int t^{2}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{4}\mathrm{d}t with \frac{t^{5}}{5}. Multiply 75 times \frac{t^{5}}{5}.

$$15t^{5}+24\int t^{2}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{2}\mathrm{d}t with \frac{t^{3}}{3}. Multiply 24 times \frac{t^{3}}{3}.

$$15t^{5}+8t^{3}$$

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.

$$15\times 3^{5}+8\times 3^{3}-\left(15\times 0^{5}+8\times 0^{3}\right)$$

Simplify.

$$3861$$