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INTEGRATION

AP CALCULUS!. INTEGRATION. FINDING AREA BETWEEN TWO CURVES AND THE VOLUME OF A SOLID. 2009 FRQ #4. Let R be the region in the first quadrant enclosed by the graphs of y=2x and y= x 2 , as shown in the figure. Find the area of R.

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INTEGRATION

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  1. AP CALCULUS! INTEGRATION FINDING AREA BETWEEN TWO CURVES AND THE VOLUME OF A SOLID

  2. 2009 FRQ #4 Let R be the region in the first quadrant enclosed by the graphs of y=2x and y= x2 , as shown in the figure. • Find the area of R. • The region R is the base of a solid. For this solid, at each x the cross section perpendicular to the x-axis has area A(x)=sin(π/2)x). Find the volume of the solid. (c) Another solid has the same base R. For this solid, the cross sections perpendicular to the y-axis are squares. Write, but do not evaluate, an integral expression for the volume of the solid.

  3. PART A: Find the area of R. Area= ∫(top function- bottom function) dx Top function-- y=2x Bottom function -- y=x2 Intersection at x= 0 and x=2 Equation: Solve: Integrate : or 1.333 So… you’re probably like… But let’s do this, yea?

  4. PART A Continued Solving for area graphically: Graph function y=2x-x2 lower limit: x=0 Calculate for area (∫f(x)dx) upper limit: x=2

  5. PART B: The region R is the base of a solid. For this solid, at each x the cross section perpendicular to the x-axis has area A(x)=sin(π/2)x). Find the volume of the solid. Volume= ∫Area dx Area= A(x)=sin(π/2)x) Functions bounding region R intersect at x= 0 and x=2 (sin(π/2)x)dx Volume = - 2/π cos(π/2)x) u= π/2 x du= π/2 dxdx= 2/π du = - 2/π[cos(π) – cos (0)] dx= 2/π du Anti-deriv of sinx = -cosx= - 2/π[-1-1] = 4/π or 1.273 We got this!

  6. PART B Continued Solving for volume graphically: Graph y=sin(π/2)x) lower limit: x=0 upper limit: x=2 Calculate for area (∫f(x)dx) Have to use the calculator?

  7. PART C: Another solid has the same base R. For this solid, the cross sections perpendicular to the y-axis are squares. Write, but do not evaluate, an integral expression for the volume of the solid. • Since the cross sections are perpendicular to the • y-axis… Rearrange the equations y=2x and y=x2 x=y/2 and x= √ ̅y Intersection of the two functions at y=0 and y=4 Volume= ∫Area dy Area= (right function-left function)2 =( √ ̅y – (y/2))2 Perpendicular to the y-axis…

  8. We’re DONE!

  9. http://apcentral.collegeboard.com/apc/public/repository/ap09_ab_sgs.pdfhttp://apcentral.collegeboard.com/apc/public/repository/ap09_ab_sgs.pdf http://i985.photobucket.com/albums/ae335/BlueMaxx86/tumblr_lc9ge9dK1R1qzado8o1_500.gif http://26.media.tumblr.com/tumblr_ljgd3kgyOi1qbp4vao1_500.gif http://24.media.tumblr.com/tumblr_ldqcp8Lmit1qbdzeqo1_500.gif http://media.tumblr.com/tumblr_ljux3ubA9J1qcgb8s.gif http://24.media.tumblr.com/tumblr_lk9qs5JbDV1qgfnv2o1_400.gif http://data.whicdn.com/images/3990503/tumblr_l9263obx841qbjwplo1_500_thumb.png?1285033365 Citations

  10. Well then… goodbye

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