# Thread: area of the region

1. ## area of the region

Decide whether to integrate with respect to x or y. Then find the area of the region

$y = 7 \cos x, y = (6 sec(x))^2, x = -\pi / 4, x = \pi / 4$

i used the formula: $A = \int_a^b(f(x)-g(x)) dx$

$A = \int_a^b ((6 sec(x))^2 - 7 cosx)dx$

$a = -\pi / 4 , b = -\pi / 4$

not sure what to do next

2. Originally Posted by viet
$A = \int_a^b ((6 sec(x))^2 - 7 cosx)dx$

$a = -\pi / 4 , b = -\pi / 4$
But $\int_a^bf(x)\,dx = 0$ if $a = b$ by the fundamental theorem of calculus.
If you change it to this it should work.
$A = \int_{-\frac{\Pi}{4}}^{\frac{\Pi}{4}}((6 sec\,(x))^2 - 7 cos\,(x))\,dx$
You could also exploit the symmetry to get
$A = 2\int_0^{\frac{\Pi}{4}}((6 sec\,(x))^2 - 7 cos\,(x))\,dx$

3. Originally Posted by fgn
If you change it to this it should work.
$A = \int_{-\frac{\Pi}{4}}^{\frac{\Pi}{4}}((6 sec\,(x))^2 - 7 cos\,(x))\,dx$
i understand that part, but not sure how to get the answer from here

4. Viet isn't the only one looking for help on this particular problem. I know a couple of others are as well. Can anyone help us through this problem step by step?

5. Originally Posted by viet
i understand that part, but not sure how to get the answer from here
So far so good!

It's good if you try to learn the derivatives of the elementary functions.
$\frac{d}{dx}(tan\,(x)) = \frac{sin^{\,2}(x) + cos^{\,2}(x)}{cos^{\,2}(x)} = sec^{\,2}(x) = tan^{\,2}(x) + 1$

From this you get $\int sec^{\,2}(x)\,dx = tan\,(x) + C$

$\int((6 sec\,(x))^2 - 7 cos\,(x))\,dx = 36tan\,(x) - 7sin\,(x) + C$

I hope you can take it from here.