Solve ∫ (from 3 to 4) of 3(4t)^2dt?

$$\int _ { 3 } ^ { 4 } 3 ( 4 t ) ^ { 2 } d t ?$$

Evaluate

$$592$$

Solution Steps

Expand \left(4t\right)^{2}.

$$\int _{3}^{4}3\times 4^{2}t^{2}\mathrm{d}t$$

Calculate 4 to the power of 2 and get 16.

$$\int _{3}^{4}3\times 16t^{2}\mathrm{d}t$$

Multiply 3 and 16 to get 48.

$$\int _{3}^{4}48t^{2}\mathrm{d}t$$

Evaluate the indefinite integral first.

$$\int 48t^{2}\mathrm{d}t$$

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

$$48\int t^{2}\mathrm{d}t$$

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}. Multiply 48 times \frac{t^{3}}{3}.

$$16t^{3}$$

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.

$$16\times 4^{3}-16\times 3^{3}$$

Simplify.

$$592$$