Key to "Use the Comparison Theorem to determine whether the..."

Rui Baldwin

Answered question

2021-10-23

Use the Comparison Theorem to determine whether the integral is convergent or divergent. integral 0 to infinity

\frac{x}{x^{3}} + 1 d x

pierretteA

Skilled2021-10-24Added 102 answers

\int_{0}^{\infty} \frac{x}{x^{3} + 1} d x \leq \int_{0}^{1} \frac{x}{x^{3} + 1} d x + \int_{1}^{\infty} \frac{x}{x^{3} + 0} d x

\int_{0}^{\infty} \frac{x}{x^{3} + 1} d x \leq \int_{0}^{1} \frac{x}{x^{3} + 1} d x + \int_{1}^{\infty} \frac{1}{x^{2}} d x

\int_{0}^{\infty} \frac{x}{x^{3} + 1} d x \leq \int_{0}^{1} \frac{x}{x^{3} + 1} d x \pm \frac{1}{x}]_{1}^{\infty}

\int_{0}^{\infty} \frac{x}{x^{3} + 1} d x \leq \int_{0}^{1} \frac{x}{x^{3} + 1} d x + [- \frac{1}{\infty} + \frac{1}{1}]

\int_{0}^{\infty} \frac{x}{x^{3} + 1} d x \leq \int_{0}^{1} \frac{x}{x^{3} + 1} d x + 1

Note that the RHS is finite, because

\frac{x}{x^{3} + 1}

is finite on the interval [0,1]
Hence LHS is finite because it is less than RHS
Result: Converges

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?