Solve I ( x ) = ∫ (from 0 to x) of (t^2-24t+143)dt

$$I ( x ) = \int _ { 0 } ^ { x } ( t ^ { 2 } - 24 t + 143 ) d t$$

Evaluate

$$\frac{x\left(x^{2}-36x+429\right)}{3}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int t^{2}-24t+143\mathrm{d}t$$

Integrate the sum term by term.

$$\int t^{2}\mathrm{d}t+\int -24t\mathrm{d}t+\int 143\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int t^{2}\mathrm{d}t-24\int t\mathrm{d}t+\int 143\mathrm{d}t$$

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}.

$$\frac{t^{3}}{3}-24\int t\mathrm{d}t+\int 143\mathrm{d}t$$

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 -24 times \frac{t^{2}}{2}.

$$\frac{t^{3}}{3}-12t^{2}+\int 143\mathrm{d}t$$

Find the integral of 143 using the table of common integrals rule \int a\mathrm{d}t=at.

$$\frac{t^{3}}{3}-12t^{2}+143t$$

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.

$$\frac{x^{3}}{3}-12x^{2}+143x-\left(\frac{0^{3}}{3}-12\times 0^{2}+143\times 0\right)$$

Simplify.

$$\frac{x\left(x^{2}-36x+429\right)}{3}$$

Differentiate w.r.t. x

$$\left(x-13\right)\left(x-11\right)$$