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Chapter 16 Solutions
CALCULUS: EARLY TRANSCENDENTALS (LCPO)
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- A soda can has a volume of 25 cubic inches. Let x denote its radius and h its height, both in inches. a. Using the fact that the volume of the can is 25 cubic inches, express h in terms of x. b. Express the total surface area S of the can in terms of x.arrow_forwardChange of variables Use the change of variables y4 = x² – 4 to evaluate the integral x Vx? – 4 dx.arrow_forwardTRANSFER TRAN SFER ACTIVITY 2: INTEGRATION THROUGH SUBSTITUTION Direction: Evaluate the following integrals. 1. S dx Vx 2. S dxarrow_forward
- Q-1/ Evaluate the integral x?e*y dA y- 1 (2, 1) D D (0. 0)arrow_forward2e dx 2x +3 Evaluate the integralarrow_forwardSET-UP ONLY the definite integral that will give the area of the regionbounded by y = x + 1 and x = 1 − y2(a) using x as the variable of integration and(b) using y as the variable of integration.arrow_forward
- 13 2x To find the definite integral dx, apply the basic integration rule. X + 16 Let u = = x< + 16. Differentiate u in terms of x, du = 2 2 x dx. Step 2 Rewrite the integration in terms of u. 2x x= 3 du x² + 16 Jo Submit Skip (you cannot come back)arrow_forwardEvaluate the definite integral. Use the integration capabilities of a graphing utility to verify your result. 2x dx Vx2 + 16 Step 1 2x To find the definite integral Vx + dx, apply the basic integration rule. + 16 Let u = x² + 16. Differentiate u in terms of x, du = A 2 x dx. Step 2 Rewrite the integration in terms of u. 2x x = 3 du V x2 + 16 = xp Step 3 Apply the power rule of integration. rx = 3 x = 3 du = du +1 1x = 3 +1Jo 1x = 3arrow_forwardEvaluate the definite integral. Use the integration capabilities of a graphing utility to verify your result. 2x dx + 64 Step 1 6 2x To find the definite integral dx, apply the basic integration rule. + 64 Let u = x- + 64. Differentiate u in terms of x, du = 4 X 2 x dx. Step 2 Rewrite the integration in terms of u. 9. 2x Xx=6 dx = du x + 64arrow_forward
- Evaluate the definite integral. Use the integration capabilities of a graphing utility to verify your result. 2x dx x2 + 16 Step 1 2x To find the definite integral dx, apply the basic integration rule. + 16 Let u = x2 + 16. Differentiate u in terms of x, du = 2 2 x dx. Step 2 Rewrite the integration in terms of u. 2x du dx = V x2 + 16arrow_forwardEvaluate the definite integral. Use the integration capabilities of a graphing utility to verify your result. 2x dx +16 Step 1 2x To find the definite integral dx, apply the basic integration rule. x² + 16 Let u = x + 16. Differentiate u in terms of x, du = 2 2 x dx. Step 2 Rewrite the integration in terms of u. x 3 1 1 du 2x x +16 Vu Step 3 Apply the power rule of integration. X= 3 x = 3 -1/2 -1/2 du du = -1/2 -1/2 + 17x = 3arrow_forwardThe integrand of the definite integral is a difference of two functions. 5 - X dx 15 15 Sketch the graph of each function and shade the region whose area is represented by the integral. 6. 4 y y 2- 2 4 6. 4 4 y y 2- 2 8.arrow_forward
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