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University of OttawaMAT 1332 First Midterm Exam | download for quality grades

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uestion 1. [4 points] Suppose that a tree trunk is 1 m high, and that its radius at height x m is r(x) = e x 2 p 1 + (e x) 2 m. Find its volume. Solution: The integral that we need to compute ... (to get the volume) is: I = R 1 0 π(r(x))2dx = π R 1 0 e x 1+(e x) 2 dx. Use the following SUB: u = e x , and notice that du dx = e x , or du = e xdx. About limits of integration: when x = 0, one has that u = e 0 = 1; when x = 1, one has that u = e 1 = e. Our integral becomes now: I = π R e 1 du 1+u2 = π{arctan(u)| e 1} = π{arctan(e) − arctan(1)} = π{arctan(e) − π 2 }. 2 Downloaded by Sam Yee ([email protected]) lOMoARcPSD|3162150 Question 2. [3 points] Solve the separable differential equation dy dt = 16tet 2+1 y with initial condition y(0) = 2√ e. Solution: This is a Separable Differential equation. Separate: ydy = 16tet 2+1dt. Integrate: R ydy = R 16tet 2+1dt. Compute as follows: (using the power rule and a SUB: u = t 2 + 1, so du dt = 2t) y 2 2 = 16 R e u du 2 = 8 R e udu = 8e u + c = 8e t 2+1 + c, where c is a number. The initial condition says: 2√ e = y(0), so 4e = y(0)2 , hence 2e = (y(0))2 2 = 8e 0+1 + c, thus c = 2e − 8e = −6e. From y 2 2 = 8e t 2+1 − 6e, one has y = ± √ 16e t 2+1 − 12e. Since the Initial Condition is Positive, it follows that the solution is ONLY: y = √ 16e t 2+1 − 12e. 3 Downloaded by Sam Yee ([email protected]) lOMoARcPSD|3162150 Question 3. [5 points] Find the indefinite integral Z 2x 3 − 18x 2 + 39x + 1 x 2 − 9x + 20 dx. SOLUTION: By LONG DIVISION one has that 2x 3−18x 2+39x+1 x2−9x+20 = 2x + −x+1 x2−9x+20 , which can be written as: 2x + −x+1 (x−4)(x−5) . For the second term we use Partial Fractions: −x+1 (x−4)(x−5) = A x−4 + B x−5 , hence −x + 1 = A(x − 5) + B(x − 4) = x(A + B) − 5A − 4B. Hence A + B = −1, and −5A − 4B = 1, Thus A = −1 − B and −5(−1 − B) − 4B = 1. We get B = −4 and A = 3. Our integral becomes: x 2 + 3 |x−4| + −4 |x−5| + c, c a number [Show More]

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