# Grading 3 points for δ x 4 points for x i 4 points

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Grading: +3 points for Δ x , +4 points for x i , +4 points for the form of Simp- son’s Rule, +4 points for substitution. Full credit was given if a different value of n was used, without making any mistakes. Grading for common mistakes: - 2 points for x 0 = 0 ; - 3 points for x 0 = 0 and x n = 1 . 5
C. HECKMAN MAT 266 Solutions, 1:40 Class Test 1 (1) [15 points] Do a trigonometric substitution on the following integral. That is, make a substitution which gets rid of the radical signs and all the x ’s. You may stop when you have gotten rid of these things; you do not need to evaluate the resulting integral. 1 + ln ( 2 x 2 - 9 + 5 x ) e 2 x cos ( 3 x - x 2 - 9 ) dx Solution: Because of the presence of x 2 - 9 = x 2 - 3 2 , the proper trigono- metric substitution is x = 3 sec θ . This means dx = 3 sec θ tan θ dθ and x 2 - 9 = 3 tan θ . Substituting these into the integral produces 1 + ln ( 2 · 3 tan θ + 5 · 3 sec θ ) e 2 · 3 sec θ cos ( 3 · 3 sec θ - 3 tan θ ) · 3 sec θ tan θ dθ. Grading: +4 points for choosing the trigonometric substitution, +4 points for dx and the square root, +4 points for substituting into the function, +3 points for replacing dx . 1
(2) Find the following integrals: (a) [10 points] 7 x - 4 x 2 + 49 dx Solution: This is already a partial fraction, since the denominator has no real roots. You had to split this fraction into two pieces, do a u -substitution on the first one, and use the arc tangent formula on the second one. 7 x - 4 x 2 + 49 dx = 7 x x 2 + 49 dx - 4 1 x 2 + 49 dx = 7 · 1 2 ln( x 2 +49) - 4 · 1 7 tan - 1 x 7 + C. Grading: +4 points for splitting up the fraction, +3 points for each piece. (b) [10 points] tan 3 θ sec 7 θ dθ Solution: This is a trigonometric integral where the power of tan θ is odd, and there is at least one sec θ ; hence, you should be preparing for the u -substitution u = sec θ . This means you need to factor out a sec θ tan θ and convert the rest of the function into sec θ ’s, using the identity sec 2 θ = 1 + tan 2 θ : tan 3 θ sec 7 θ dθ = tan 2 θ sec 6 θ · sec θ tan θ dθ = (sec 2 θ - 1) sec 6 θ · sec θ tan θ dθ = ( u 2 - 1) · u 6 du u = sec θ du = sec θ tan θ du = sec θ tan θ dθ = u 8 - u 6 du = u 9 9 - u 7 7 + C = sec 9 θ 9 - sec 7 θ 7 + C. Grading: +3 points for the idea, +2 points for converting to sec θ ’s, +2 points for multiplying out the polynomial, +3 points for finishing up. = 2
(c) [10 points] x 2 e - 4 x dx Solution: Integration by parts, twice, then a u -substitution: x 2 e - 4 x dx = x 2 · e - 4 x - 4 - 2 x · e - 4 x - 4 dx u = x 2 v = e - 4 x u = 2 x v = e - 4 x - 4 = - x 2 e - 4 x 4 + 1 2 xe - 4 x dx = - x 2 e - 4 x 4 + 1 2 x · e - 4 x - 4 - 1 · e - 4 x - 4 dx u = x v = e - 4 x u = 1 v = e - 4 x - 4 = - x 2 e - 4 x 4 + 1 2 - xe - 4 x 4 - e - 4 x 16 + C. Grading: +4 points for finding the first integration by parts, +2 points for doing it, +2 + 2 points for finding and doing the next integration by parts.