Solve ∫ (from 0 to 3) of t^2-14t+49dt

$$\int _ { 0 } ^ { 3 } t ^ { 2 } - 14 t + 49 d t$$

Evaluate

$$93$$

Solution Steps

Evaluate the indefinite integral first.

$$\int t^{2}-14t+49\mathrm{d}t$$

Integrate the sum term by term.

$$\int t^{2}\mathrm{d}t+\int -14t\mathrm{d}t+\int 49\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int t^{2}\mathrm{d}t-14\int t\mathrm{d}t+\int 49\mathrm{d}t$$

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}.

$$\frac{t^{3}}{3}-14\int t\mathrm{d}t+\int 49\mathrm{d}t$$

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 -14 times \frac{t^{2}}{2}.

$$\frac{t^{3}}{3}-7t^{2}+\int 49\mathrm{d}t$$

Find the integral of 49 using the table of common integrals rule \int a\mathrm{d}t=at.

$$\frac{t^{3}}{3}-7t^{2}+49t$$

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.

$$\frac{3^{3}}{3}-7\times 3^{2}+49\times 3-\left(\frac{0^{3}}{3}-7\times 0^{2}+49\times 0\right)$$

Simplify.

$$93$$