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How to solve these difficult AP Calculus problems?

  • I took a practice AP Calculus AB test from my Kaplan study guide (2007) and couldn't figure out these two explanations (some study guide...) These are from the free response section. How to solve 160=integral from 0 to S: 10 + 6sin((t^2)/3)dt for S. In calculator speak that's 160 = fint(10 + 6sin((X^2)/3)), X, 0, S) Somehow, I should be able to solve this using a calculator. The best way that I could this was by plotting 10 + 6sin((X^2)/3)) and using the integration option with different intervals from 0 to a number, gradually narrowing down the interval. On the TI-84 Plus, ie., I would plot the function, press CALC (2nd and TRACE), and press 7 (integral). This would allow me to set a left and right bound, 0 and S, wherein I would have to gradually narrow down S till the it equaled 160. That takes a lot of time. The answer the book gave was S=15.2982 Is there another way to solve this, especially analytically? Second, on another problem, without a calculator, I have to solve π* integral from 0 to 1: X- (1 -cos(πX/2))^2 dX In calculator speak that's π fint (X- (1 -cos(πX/2))^2, X, 0, 1) except I can't use a calculator... There is probably a trigonometric identity involved... The answer the book gave was 0.8584 Thanks in advance!

  • Answer:

    X-1+2*cos(pi*x/2)-cos^2(pi*x/2) .5x^2-x+4/pi*sin(pi*x/2)-integral cos^2(pi*x/2) integral cos^2(pi*x/2)=integral (1+cospi*x)/2=x/2+(sinpix)/(pi*2) answer=.5x^2-1.5x+4/pi*sin(pi*x/2)-(si... from zero to one yields -1+4/pi= multply this by pi to get your ans also, you may want to get a ti 89 since integration is a breeze on there.

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X-1+2*cos(pi*x/2)-cos^2(pi*x/2) .5x^2-x+4/pi*sin(pi*x/2)-integral cos^2(pi*x/2) integral cos^2(pi*x/2)=integral (1+cospi*x)/2=x/2+(sinpix)/(pi*2) answer=.5x^2-1.5x+4/pi*sin(pi*x/2)-(si… from zero to one yields -1+4/pi= multply this by pi to get your ans also, you may want to get a ti 89 since integration is a breeze on there.

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