Solve ∫ (from 0 to 1) of (6t-3t)^2dt

$$\int _ { 0 } ^ { 1 } ( 6 t - 3 t ) ^ { 2 } d t$$

Evaluate

$$3$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \left(6t-3t\right)^{2}\mathrm{d}t$$

Factor out the constant using \int af\left(t\right)\mathrm{d}t=a\int f\left(t\right)\mathrm{d}t.

$$\left(6-3\right)^{2}\int t^{2}\mathrm{d}t$$

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}.

$$\left(6-3\right)^{2}\times \frac{t^{3}}{3}$$

Simplify.

$$3t^{3}$$

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.

$$3\times 1^{3}-3\times 0^{3}$$

Simplify.

$$3$$