Solve ∫ (from 0 to 2) of (30+10t-5/2t)dt

$$\int _ { 0 } ^ { 2 } ( 30 + 10 t - \frac { 5 } { 2 } t ) d t$$

Evaluate

$$75$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 30+10t-\frac{5t}{2}\mathrm{d}t$$

Integrate the sum term by term.

$$\int 30\mathrm{d}t+\int 10t\mathrm{d}t+\int -\frac{5t}{2}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int 30\mathrm{d}t+10\int t\mathrm{d}t-\frac{5\int t\mathrm{d}t}{2}$$

Find the integral of 30 using the table of common integrals rule \int a\mathrm{d}t=at.

$$30t+10\int t\mathrm{d}t-\frac{5\int t\mathrm{d}t}{2}$$

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 10 times \frac{t^{2}}{2}.

$$30t+5t^{2}-\frac{5\int t\mathrm{d}t}{2}$$

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 -\frac{5}{2} times \frac{t^{2}}{2}.

$$30t+5t^{2}-\frac{5t^{2}}{4}$$

Simplify.

$$30t+\frac{15t^{2}}{4}$$

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.

$$30\times 2+\frac{15}{4}\times 2^{2}-\left(30\times 0+\frac{15}{4}\times 0^{2}\right)$$

Simplify.

$$75$$