Solve ∫ (from 3 to 4) of (-40+10t)(95-40t+5t^2)dt

$$\int _ { 3 } ^ { 4 } ( - 40 + 10 t ) ( 95 - 40 t + 5 t ^ { 2 } ) d t$$

Evaluate

$$-\frac{175}{2}=-87.5$$

Solution Steps

Use the distributive property to multiply -40+10t by 95-40t+5t^{2} and combine like terms.

$$\int _{3}^{4}-3800+2550t-600t^{2}+50t^{3}\mathrm{d}t$$

Evaluate the indefinite integral first.

$$\int -3800+2550t-600t^{2}+50t^{3}\mathrm{d}t$$

Integrate the sum term by term.

$$\int -3800\mathrm{d}t+\int 2550t\mathrm{d}t+\int -600t^{2}\mathrm{d}t+\int 50t^{3}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int -3800\mathrm{d}t+2550\int t\mathrm{d}t-600\int t^{2}\mathrm{d}t+50\int t^{3}\mathrm{d}t$$

Find the integral of -3800 using the table of common integrals rule \int a\mathrm{d}t=at.

$$-3800t+2550\int t\mathrm{d}t-600\int t^{2}\mathrm{d}t+50\int t^{3}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t\mathrm{d}t with \frac{t^{2}}{2}. Multiply 2550 times \frac{t^{2}}{2}.

$$-3800t+1275t^{2}-600\int t^{2}\mathrm{d}t+50\int t^{3}\mathrm{d}t$$

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 -600 times \frac{t^{3}}{3}.

$$-3800t+1275t^{2}-200t^{3}+50\int t^{3}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{3}\mathrm{d}t with \frac{t^{4}}{4}. Multiply 50 times \frac{t^{4}}{4}.

$$-3800t+1275t^{2}-200t^{3}+\frac{25t^{4}}{2}$$

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.

$$1275\times 4^{2}-3800\times 4-200\times 4^{3}+\frac{25}{2}\times 4^{4}-\left(1275\times 3^{2}-3800\times 3-200\times 3^{3}+\frac{25}{2}\times 3^{4}\right)$$

Simplify.

$$-\frac{175}{2}$$