Solve ∫ (from 0 to 2) of 4texp(2t)dt?

$$\int _ { 0 } ^ { 2 } 4 t \exp ( 2 t ) d t ?$$

Evaluate

$$\frac{64epx}{3}$$

Solution Steps

Multiply t and t to get t^{2}.

$$\int _{0}^{2}4t^{2}exp\times 2\mathrm{d}t$$

Multiply 4 and 2 to get 8.

$$\int _{0}^{2}8t^{2}exp\mathrm{d}t$$

Evaluate the indefinite integral first.

$$\int 8t^{2}exp\mathrm{d}t$$

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

$$8exp\int t^{2}\mathrm{d}t$$

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}.

$$8exp\times \frac{t^{3}}{3}$$

Simplify.

$$\frac{8expt^{3}}{3}$$

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.

$$\frac{8}{3}exp\times 2^{3}-\frac{8}{3}exp\times 0^{3}$$

Simplify.

$$\frac{64exp}{3}$$

Differentiate w.r.t. x

$$\frac{64ep}{3}$$