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The area of the shaded part in the figure could be represented as: 7FY y=5 y=1+x² L L -3 -2 -1 5 A = L²(x² (x²+4) dx 0 A = L²(x²-4) dx 0 2 ^^= [(4-x²) de A 0 ^= S²(17²-4) de A dx 0 6 5 4 3 32 2 1 1 2 3

Find the length of the curve. y = 3x 3/2 from x = 0 to x = 5 0 1 O 335 3 O 335 243 O 335 162 ola

A solid is formed by revolving the shaded region about the line y = 1 in the graph below. y=5-x² y=2 - y=1 2 -1 1. What is the outer radius R(x)? 2. What is the inner radius r(x)? 3. What is the volume? 1

Evaluate S 1. What is a²? 2. What is a? 3. What is u²? 4. What is u? 5. What is du? 6. What is the result of integration? x5 100+ x¹2 -dx

Let R be the region enclosed by the x-axis, 1≤ x≤ 3, and the curve y ex. Set up the volume by rotating the R about the x-axis. y = e² V=n ex dx ° v= f²c²d² ex dx 1 3 Se V=A e2x dx 1 ?v=f0²d0 V= e 2x dx R 3 x

Find the area of the shaded region about x-axis. Find Areal Area 2 correctly to one decimal place. y 60 50 40 30 y = 4x+16 20 -2 -1 0 1 y=2x2+10 2 3 5 Area 3 and the total Area Give your answers either in fraction form, a/b or

Set up the correct volume by rotating the region bounded by two parabolas y = x² + 1 and y = 3 - x² about the x-axis. YA y=x² + 1 \y=3=x² -1 0 1 V=₁²₁ (2x² - 2)dx -1 °V= π² (2x²¹ - 8)dx -1 O °V= π (8x² - 8)dx ⒸV-t (2xª-4x²-8)dx = X

Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. y = cosx, 0≤ x ≤ π/4; x-axis π/4 2TT f sinx V1 + cosx dx 7/4 2TT cosx V1 + sin2x dx Į π/4 7/4 sinx Vì + cosx dx cosx Vit sin²x dx

Calculate the surface area generated by revolving the curve O 2000 6366.4 2026.5 π O 2026.5 y=-x³, from x = 0 to x = 3 about the x-axis. 3

Evaluate the integral. x5 finx³dx 1 (In x5)² + C (In x5)² + C ¡ (In x5)² + C + C O о 10 In x5

Determine the volume of the shaded region rotated about the x-axis in the figure below. 8 y=2x y = −2x2 + 8x 1 1. What is the outer radius R(x)? 2. What is the inner radius r(x)? 3. What is the volume? 2 3