Solving Improper Integrals: \int_{3}^{6}, \lim_{t\rightarrow\infty}

[tandoorichicken]

How do I do
$$\int_{3}^{6} (5-x)^{\frac{-1}{3}} \,dx$$
?

I also need to know how to do
$$\lim_{t\rightarrow\infty} \int_{-t}^{t} \frac{x}{x^2 + 1} \,dx$$

 

[ahrkron]

You could do both by substitution. Try the second one first. Define
$$u=x^2+1$$.
What is $$du$$?

However, there is a hidden trick on the second one... draw the integrand as a function of x... can you tell its area from -t to t "by eye"?

 

[M98Ranger]

To solve the first integral, we can use the substitution method. Let u = 5-x, then du = -dx. When x = 3, u = 2 and when x = 6, u = -1. Therefore, the integral becomes:

\int_{3}^{6} (5-x)^{\frac{-1}{3}} \,dx = -3 \int_{2}^{-1} u^{\frac{-1}{3}} \,du

Using the power rule for integration, we get:

-3 \int_{2}^{-1} u^{\frac{-1}{3}} \,du = -3 \left[\frac{3u^{\frac{2}{3}}}{2}\right]_{2}^{-1} = -\frac{9}{2} \left[\left(\frac{1}{2}\right)^{\frac{2}{3}} - \left(\frac{1}{5}\right)^{\frac{2}{3}}\right] = \frac{9}{10}

To solve the second integral, we can use the limit comparison test. First, let's rewrite the integral as:

\int_{-t}^{t} \frac{x}{x^2 + 1} \,dx = \int_{-t}^{t} \frac{x}{(x + 1)(x - 1)} \,dx

Now, we can use the limit comparison test by comparing it to the integral \int_{-t}^{t} \frac{1}{x-1} \,dx which is a known improper integral with a known solution. We get:

\lim_{t\rightarrow\infty} \frac{\int_{-t}^{t} \frac{x}{x^2 + 1} \,dx}{\int_{-t}^{t} \frac{1}{x-1} \,dx} = \lim_{t\rightarrow\infty} \frac{\frac{1}{2}\ln(t+1) - \frac{1}{2}\ln(t-1)}{\ln(t-1)} = \lim_{t\rightarrow\infty} \frac{\frac{1}{2}\ln\left(\frac{t+1}{t-1}\right)}{\ln(t-1)} = \frac{1}{2}