Evaluating Integrals

**RezMan** · June 6th 2011, 02:50 AM

hi guys can you please help me with this:

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)?

**TKHunny** · June 6th 2011, 03:39 AM

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

**Prove It** · June 6th 2011, 04:18 AM

Originally Posted by RezMan

hi guys can you please help me with this:

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.

**HallsofIvy** · June 6th 2011, 06:43 AM

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,
$\int_a^b f(x)dx= -\int_b^a f(x)dx$

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

**RezMan** · June 6th 2011, 11:22 AM

okay so I get this:

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

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

?

**TKHunny** · June 6th 2011, 03:07 PM

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.

**HallsofIvy** · June 7th 2011, 09:29 AM

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.

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