MATH 105 Fall 2011 Sample Midterm 2 Solutions - Math 105 Practice Midterm 2 1 Compute the following a d dx 2 1 dt x 1 t3 x d 1 Solution We rewrite this

MATH 105 Fall 2011 Sample Midterm 2 Solutions - Math 105...

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Math 105 - Practice Midterm 2 1 Compute the following: a d dx R 2 x 1 1+ t 3 dt Solution: We rewrite this as - d dx R x 2 1 1+ t 3 dt and apply the Funda- mental Theorem of Calculus to yield - 1 1+ x 3 . b R -∞ xdx Solution: This needs to be split into two improper integrals because it ranges over ( -∞ , ) . If the integral converges, I = R 0 -∞ xdx + R 0 xdx . Both must converge for the integral to converge. Note that R 0 xdx = lim R →∞ R R 0 xdx = lim R →∞ x 2 2 | R 0 = . Thus I diverges and we don’t even need to consider the other integral. c R 0 dx x 1 / 2 + x 3 / 2 Solution: This integral is improper for two reasons: the vertical asymptote at x = 0 and the infinite range of integration. We split it into R 1 0 dx x 1 / 2 + x 3 / 2 + R 1 dx x 1 / 2 + x 3 / 2 . Both must converge for the integral to converge. Making a substitution u = x 1 / 2 with du = 1 2 x 1 / 2 dx so dx = 2 udu in both integrals, we have R dx x 1 / 2 + x 3 / 2 = R 2 1+ u 2 du = 2 arctan u + C = 2 arctan x + C . Thus, R 0 dx x 1 / 2 + x 3 / 2 = lim t 0 + 2 arctan x | 1 t +lim R →∞ 2 arctan x | R 1 = 2( π/ 4 - 0) + 2( π/ 2 - π/ 4) = π. d R dy 1+cos(4 y ) Solution: We write 1 + cos(4 y ) = 2 cos 2 (2 y ) . Thus, R dy 1+cos(4 y ) = R 1 2 sec 2 (2 y ) = 1 4 tan(2 y ) + C. e R (ln( x 8 )) 5 dx Solution: First of all we pull out the 8 to find R (8 ln x ) 5 dx = 8 5 R ln 5 xdx. This will involve multiple integration-by-parts. We’ll first note that R ln n xdx = R ln n x 1 dx so that if f = ln n x with df = n (ln n - 1 x ) 1 x and dg = dx with g = x then R ln n xdx = x ln n x - n R ln n - 1 xdx. Life is good now: 8 5 R ln 5 xdx = 8 5 ( x ln 5 x - 5 R ln 4 xdx ) = 8 5 ( x ln 5 x - 5( x ln 4 x - R 4 ln 3 xdx )) = 8 5 ( x ln 5 x - 5 x ln 4 x +20( R ln 3 xdx )) = |{z} guess pattern 8 5 ( x ln 5 x - 5 x ln 4 x +5 × 4 x ln 3 x - 5 × 4 × 3 x ln 2 x +5 × 4 × 3 × 2 ln x - 5 × 4 × 3 × 2 × 1 x ) = 8 5 ( x ln 5 x - 5 x ln 4 x + 20 x ln 3 x - 60 x ln 2 x + 120 x ln x - 120 x ) + C . This result can be verified by taking a derivative. 1
f R (tan 2 t sec t - tan 3 t sec t ) dt Solution: We split the integral into two, as they require different strategies. I = R tan 2 t sec tdt = R (sec 2 t - 1) sec tdt = R sec 3 t - R sec tdt.

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