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Grading:+3points forΔx,+4points forxi,+4points for the form of Simp-son’s Rule,+4points for substitution. Full credit was given if a different valueofnwas used, without making any mistakes.Grading for common mistakes:-2points forx0= 0;-3points forx0= 0andxn= 1.5
C. HECKMANMAT 266Solutions, 1:40 ClassTest 1(1) [15 points] Do a trigonometric substitution on the following integral. That is, makea substitution which gets rid of the radical signs and all thex’s. You may stop whenyou have gotten rid of these things; you do not need to evaluate the resulting integral.1 + ln(2√x2-9 + 5x)e2xcos(3x-√x2-9)dxSolution:Because of the presence ofx2-9 =x2-32, the proper trigono-metric substitution isx= 3 secθ. This meansdx= 3 secθtanθ dθandx2-9 =3 tanθ. Substituting these into the integral produces1 + ln(2·3 tanθ+ 5·3 secθ)e2·3 secθcos(3·3 secθ-3 tanθ)·3 secθtanθ dθ.Grading:+4points for choosing the trigonometric substitution,+4points fordxand the square root,+4points for substituting into the function,+3points forreplacingdx.1
(2) Find the following integrals:(a) [10 points]7x-4x2+ 49dxSolution:This is already a partial fraction, since the denominator has no realroots. You had to split this fraction into two pieces, do au-substitution on thefirst one, and use the arc tangent formula on the second one.7x-4x2+ 49dx= 7xx2+ 49dx-41x2+ 49dx= 7·12ln(x2+49)-4·17tan-1x7+C.Grading:+4points for splitting up the fraction,+3points for each piece.(b) [10 points]tan3θsec7θ dθSolution:This is a trigonometric integral where the power oftanθis odd, andthere is at least onesecθ; hence, you should be preparing for theu-substitutionu= secθ. This means you need to factor out asecθtanθand convert the rest ofthe function intosecθ’s, using the identitysec2θ= 1 + tan2θ:tan3θsec7θ dθ=tan2θsec6θ·secθtanθ dθ=(sec2θ-1) sec6θ·secθtanθ dθ=(u2-1)·u6duu= secθdudθ= secθtanθdu= secθtanθ dθ=u8-u6du=u99-u77+C=sec9θ9-sec7θ7+C.Grading:+3points for the idea,+2points for converting tosecθ’s,+2pointsfor multiplying out the polynomial,+3points for finishing up.=⇒2
(c) [10 points]x2e-4xdxSolution:Integration by parts, twice, then au-substitution:x2e-4xdx=x2·e-4x-4-2x·e-4x-4dxu=x2v=e-4xu= 2xv=e-4x-4=-x2e-4x4+12xe-4xdx=-x2e-4x4+12x·e-4x-4-1·e-4x-4dxu=xv=e-4xu= 1v=e-4x-4=-x2e-4x4+12-xe-4x4-e-4x16+C.Grading:+4points for finding the first integration by parts,+2points for doingit,+2 + 2points for finding and doing the next integration by parts.