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# 2015 AP Calculus AB 2a

Areas between curves.

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• Hmm. I usually just split the functions up into multiple intervals. In this case it was from 0 to 1.033 for R, and 1.033 to 2 for S. I guess with a calculator this way is much easier, but I was taught to separate the integrals. • Yeah, separating them into parts is pretty standard as well. Really, it's accomplishing the same thing because when you evaluate the integral of an absolute value function, at every point the inner function would have switched signs you would have to split the integral and evaluate the positive or negative version of whatever is inside the absolute value signs. That amounts to being the same thing you would do if you split the integral right away to solve this problem.

Still, it's important to know how to split integrals in case you have a shape that's defined by three or more functions and the interaction isn't as simple as a single intersection point.
• Let's say I want to complicate my life a little and find the areas as the sum of two separate areas R and S.
1) How do solve the equation in order to find the point of intersection? (x^4 - 6.5 * x^2 + 5 * x + 1 - e^(x^2 - 2 * x) = 0) It appears to be unsolvable.
2) Assuming problem #1 has been resolved, what is the effective way to find which function is greater? Or should I simply take an absolute value and that should do the job?
3) How do you solve the integral of (e^(x^2 - 2 * x)dx)?
4) If I do not have a calculator on the exam, does that mean I am doomed? • From my experience, yes. There is no easy way to solve an equation where: a^f(x) = g(x). The only way to do this simply is with the equation solver on a TI-84 or other calculator with similar abilities. If you plan on taking the AP Calculus exam, you should invest in a calculator. Additionally, on problem 1 in this series, you have to take the definite integral of sin(x^2) or on this question the definite integral of e^(x^2 - 2x)dx. This isn't possible without a graphing calculator or memorization of complex irregular integrals that the test designers don't expect you to know.

As for your question 2, I look at the graphs above. Perhaps this is something I can only see, but the graph of g(x) is clearly the top border of R and the bottom border of S. This may have something to do with the fact that g(x) is a degree 4 polynomial and f(x) is an exponentially based function.

Another way to explain is to take the derivative of g(x) and see that it should have 2 roots between 0 and 2 - horizontal tangents, while the exponential function has at maximum one.
• For the question 2a at , why don't we have to split up the two portions of the graph to integrate them. In section R, g(x) is above the f(x) line however in section S the reverse is true. Wouldn't this affect the answer? • Nice solution! Much easier than what I was trying to do , which as Joe pointed out was trying to figure out where f and g crossed in the interval and separate them out. I thought it kinda looked like where the inflection point was in the polynomial but that was 1.041 and actual value was 1.032832 (close). Can someone PLEASE tell me what the indefinite integral is on the exponential function , I was thinking x + x^2/2 + e ^(x^2-6) but I'm guessing there must be something more on the e term given the complex exponent. Doesnt look like a simple substitution cause if u= x^2 -6 then I cant get rid of the x. Soooo, I set u equal to the whole e term, then when integrating I get u^2/2 and when resubstituting back in, the 2 can come down from the exponent and cancel with the denominator and your left with just the original e term in the function as your integral on that term. Is that correct?? • I still don't get how he shrunk the area of both regions into the integral of | f(x)-g(x) |dx. Would it work it he still kept the same limits of integration 0 and 2 and wrote it as the integral of [g(x)-f(x)] + [f(x)-g(x)] for R + S?   