Solve ∫ 1/4(120-t)^2dt

$$\int \frac { 1 } { 4 } ( 120 - t ) ^ { 2 } d t$$

Evaluate

$$-\frac{\left(120-t\right)^{3}}{12}+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(120-t\right)^{2}.

$$\int \frac{1}{4}\left(14400-240t+t^{2}\right)\mathrm{d}t$$

Use the distributive property to multiply \frac{1}{4} by 14400-240t+t^{2}.

$$\int 3600-60t+\frac{1}{4}t^{2}\mathrm{d}t$$

Integrate the sum term by term.

$$\int 3600\mathrm{d}t+\int -60t\mathrm{d}t+\int \frac{t^{2}}{4}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int 3600\mathrm{d}t-60\int t\mathrm{d}t+\frac{\int t^{2}\mathrm{d}t}{4}$$

Find the integral of 3600 using the table of common integrals rule \int a\mathrm{d}t=at.

$$3600t-60\int t\mathrm{d}t+\frac{\int t^{2}\mathrm{d}t}{4}$$

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 -60 times \frac{t^{2}}{2}.

$$3600t-30t^{2}+\frac{\int t^{2}\mathrm{d}t}{4}$$

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 \frac{1}{4} times \frac{t^{3}}{3}.

$$3600t-30t^{2}+\frac{t^{3}}{12}$$

If F\left(t\right) is an antiderivative of f\left(t\right), then the set of all antiderivatives of f\left(t\right) is given by F\left(t\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$3600t-30t^{2}+\frac{t^{3}}{12}+С$$

Differentiate w.r.t. t

$$\frac{\left(120-t\right)^{2}}{4}$$