oldmidterms

# oldmidterms - Math 152 Sp. 10 Z. Fiedorowicz Name: EA 170...

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1) (15 points) In what follows below C and D are constants of integration. (a) If Z f ( x ) dx = 6 x x 2 + 3 + C , ﬁnd f ( x ). d dx ± 6 x x 2 + 3 ² = 6( x 2 + 3) - 6 x (2 x ) ( x 2 + 3) 2 = 18 - 6 x 2 x 4 + 6 x 2 + 9 (b) If Z x 1 g ( t ) dt = 4 x x 2 + 5 + D , ﬁnd g ( t ) and D By the fundamental theorem of calculus g ( x ) = d dx ³ 4 x x 2 + 5 ´ = 4 x 2 + 5 + 4 x x x 2 + 5 = 4( x 2 + 5) + 4 x 2 x 2 + 5 = 8 x 2 + 20 x 2 + 5 Thus g ( t ) = 8 t 2 + 20 t 2 + 5 Also plugging in x = 1 we get 0 = Z 1 1 g ( t ) dt = 4(1) 1 2 + 5 + D Thus D = - 4 6 2
2) (18 points) (a) Express the following sum sin ± - 6 π 3 + 5 π n ² 5 π n + sin ± - 6 π 3 + 10 π n ² 5 π n + sin ± - 6 π 3 + 15 π n ² 5 π n + ... + sin ± - 6 π 3 + 5 n ² 5 π n in the form Σ n i =1 a i . n X i =1 sin ± - 6 π 3 + 5 n ² 5 π n (b) Find a deﬁnite integral R b a f ( x ) dx for which the above sum is a right Riemann sum. Z b a sin( x + c ) dx where b - a = 5 π and c = - 6 π 3 - a (c) Use (b) to ﬁnd the limit of the above sum as n -→ ∞ . 2 3

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calculator will receive no credit.) (a) Z ( e 4 t - 4) 2 e 2 t dt Z ( e 4 t - 4) 2 e 2 t dt = Z e 8 t - 8 e 4 t + 16 e 2 t dt = Z ( e 6 t - 8 e 2 t + 16 e - 2 t ) dt = e 6 t 6 - 8 2 e 2 t - 16 2 e - 2 t + C (b) Z - 4 P + 2 1 - P 2 dP Z - 4 P + 2 1 - P 2 dP = - 4 Z P 1 - P 2 dP + 2 Z 1 1 - P 2 dP = - 4 Z P u du ( - 2 P ) + 2 sin - 1 ( P ) = +4 u + 2 sin - 1 ( P ) + C = +4 1 - P 2 + 2 sin - 1 ( P ) + C where we substituted u = 1 - P 2 in the ﬁrst integral on the right. 4
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## oldmidterms - Math 152 Sp. 10 Z. Fiedorowicz Name: EA 170...

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