1. ## Evaluating Integrals

If f(x)= cos^4(x)sin(x), evaluate $\int_0^\pi \! f(t) \, \mathrm{d}t.$

Would this be correct:

u=cos(x)
du=-sin(x)dx

- $\int_0^\pi \! u^4 \, \mathrm{d}u$
cos(0)=1, cos(pi)=-1

$-[\frac{u^5}{5}]^{-1}_{1}$

Is there a problem that it's decreasing from a to b? Do I need to switch something before I do F(b)-F(a)?

2. You can switch the limits around if you like. You WILL get the wrong result.

3. Originally Posted by RezMan

If f(x)= cos^4(x)sin(x), evaluate $\int_0^\pi \! f(t) \, \mathrm{d}t.$

Would this be correct:

u=cos(x)
du=-sin(x)dx

- $\int_0^\pi \! u^4 \, \mathrm{d}u$
cos(0)=1, cos(pi)=-1

$-[\frac{u^5}{5}]^{-1}_{1}$

Is there a problem that it's decreasing from a to b? Do I need to switch something before I do F(b)-F(a)?
If you switch the terminals you will need to negate the integrand.

4. The reason the limits of integration got "swapped" is that cos(x) is a decreasing function from 0 to $\pi$. And, of course, that means the derivative is negative.

As TKHunny and Prove It said,
$\int_a^b f(x)dx= -\int_b^a f(x)dx$

$\int_0^\pi cos^4(x)sin(x)dx= -\int_1^{-1} u^4du= \int_{-1}^1 u^4 du$

5. okay so I get this:

$[\frac{(1)^5}{5} - \frac{(-1)^5}{5}]^{1}_{1}$

$\frac{1}{5} - \frac{-1}{5} = \frac{2}{5}$

?

6. When you get the correct result, you should end with "finis" or something more authoritative. The question mark is just really unsatisfying.

If you remain confused, there is nothing wrong with working on the indefinite integral and simply converting back to the original variable and thus managing to use the original units.

7. Originally Posted by TKHunny
When you get the correct result, you should end with "finis" or something more authoritative. The question mark is just really unsatisfying.

If you remain confused, there is nothing wrong with working on the indefinite integral and simply converting back to the original variable and thus managing to use the original units.
Right. Instead of "?" use "!"

Yes, 2/5 is correct.