Question

\(5 - 32\)Determine whether each integral is convergent or divergent. Evaluate those that are convergent.

12. \(\int_{ - \infty }^\infty {\left( {{y^3} - 3{y^2}} \right)} dy\)

Short answer

Integral is divergent.

Step-by-step solution

Definition.

Consider the integral\(\int_{\bf{a}}^{\bf{t}} {\bf{f}} {\bf{(x)dx}}\)is for the limit with the finite number\(t \ge a\). Then, the integral is written as:

\(\int_a^\infty f (x)dx = \mathop {\lim }\limits_{t \to \infty } \int_a^t f (x)dx\)

Break integral

In order to determine if integral converges or diverges we evaluate both integrals i.e; \(\int_a^\infty f (x)dx\,\& \,\mathop {\lim }\limits_{t \to \infty } \int_a^t f (x)dx\) and see if both integrals are equal or not.

The given integral can be written as:

\(\int_{ - \infty }^\infty {\left( {{y^3} - 3{y^2}} \right)} dy = \int_{ - \infty }^0 {\left( {{y^3} - 3{y^2}} \right)} dy + \int_0^\infty {\left( {{y^3} - 3{y^2}} \right)} dy\)

Now evaluate both integrals separately & then check convergent or divergent.

Evaluate\(\int_0^\infty  {\left( {{y^3} - 3{y^2}} \right)} dy\):

Evaluate the integral as:

\(\begin{aligned}{c}\int_0^\infty {\left( {{y^3} - 3{y^2}} \right)} dy &= \mathop {\lim }\limits_{t \to \infty } \int_0^t {\left( {{y^3} - 3{y^2}} \right)} dy\\ &= \mathop {\lim }\limits_{t \to \infty } \left( {\frac{{{y^4}}}{4} - {y^3}} \right)_0^t\\ &= \mathop {\lim }\limits_{t \to \infty } \left( {{t^3}\left( {1 - \frac{t}{4}} \right)} \right)\\ &= \infty \end{aligned}\)

Evaluate\(\int_{ - \infty }^0 {\left( {{y^3} - 3{y^2}} \right)} dy\):

Evaluate the integral as:

\(\begin{aligned}{c}\int_{ - \infty }^0 {\left( {{y^3} - 3{y^2}} \right)} dy &= \mathop {\lim }\limits_{t \to - \infty } \int_t^0 {\left( {{y^3} - 3{y^2}} \right)} dy\\ &= \mathop {\lim }\limits_{t \to - \infty } \left( {\frac{{{y^4}}}{4} - {y^3}} \right)_t^0\\ &= \mathop {\lim }\limits_{t \to - \infty } \left( {\left( {{t^3} - \frac{{{t^4}}}{4}} \right)} \right)\\ &= \infty \end{aligned}\)

Since both integral are infinite so the given integral is divergent.