Sign Up For Free

**Join for FREE**,
and start talking with other members, weighing in on community polls,
and more.

Also, by registering and logging in you'll see fewer ads and pesky welcome messages (like this one!)

- Reply to threads, and start your own
- Create reports of your
**campus visits** - Share college
**photos**and**videos** **Find your dream college**, save your search and share with friends- Receive our
**monthly newsletter**

Home
/
College Discussion / College Admissions and Search / SAT and ACT Tests & Test Preparation / AP Tests Preparation

mook32
Posts: **133**Registered User Junior Member

So I was searching the internet for a breakdown of the AP Calculus AB test and I was shocked when I realized that there are 28 multiple choice questions without a calculator. Now, I can get all the multiple choice questions with calculator right, but the non-calculator section just gets to me. Does anyone else have this problem and how do you guys manage to solve the problems?

Post edited by mook32 on

## Replies to: AP Calculus Multiple Choice

1,413Registered User Senior Member4,276Registered User Senior Member133Registered User Junior Member1,413Registered User Senior Member2,741Registered User Senior MemberBtw, did the question ask you to find the exact value or just plug in the number and simplify? I'm just asking since I got 2sq(x-1)-2arctan(sq(x-1))+C as the antiderivative. (I did it by hand and verified it with the TI-89).

Oh if you want to know how i did it:

u=sq(x-1)

x=u^2+1

dx=2u du

So, plug it in: and you get int(u/(u^2+1) x 2u) du

So, i multiply it and I got int(2u^2/(u^2+1)) du. I have to separate by dividing 2u^2 by u^2+1. So, I have int(2-2/(u^2+1))du. Finally integrate and I got 2u - 2 arctan u + C. Substitute the u and you get 2sq(x-1)-2arctan(sq(x-1))+C. And you can plug in 2 and 5 to find the value from there.

And non-calculator MC shouldn't be that hard. It's just simple calculus computation. If you know the differentiation, integration, and other rules of calculus, you can do well enough to ace that portion.

223Registered User Junior Memberu = (x-1)^(1/2)

u^2 = x-1

u^2 + 1 = x

dx/du = 2u

dx = 2u du

u(2) = (2-1)^(1/2) = 1

u(5) = (5-1)^(1/2) = 2

So, changing the limits from 2, 5 to 1, 2 and substituting the expressions above, the integral becomes the integral from 1 to 2 of [u/(u^2 + 1)]*2u*du = [2u^2/(u^2 + 1)]du.

976- Member133Registered User Junior Member2,741Registered User Senior Member133Registered User Junior Member223Registered User Junior Member