Exam2_Practice_Problems_Solutions_M1620_F2019.pdf - Math...

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Math 1620Exam 2 Practice Problems SolutionsFall 20191. Evaluate the integral.(a)Z3x2+ 6x+ 2x2+ 3x+ 2dxSince the degree of the polynomial in the numerator is greater than or equal to thedegree of the polynomial in the denominator, we need to use long division, which gives3x2+ 6x+ 2x2+ 3x+ 2= 3 +-3x-4x2+ 3x+ 2.Now, on the second term, we need to use partial fraction decomposition...-3x-4x2+ 3x+ 2=-3x-4(x+ 1)(x+ 2)=Ax+ 1+Bx+ 2=A(x+ 2) +B(x+ 1)(x+ 1)(x+ 2)=⇒ -3x-4 =A(x+ 2) +B(x+ 1) =x(A+B) + (2A+B)=A+B=-3 and 2A+B=-4Subtracting first equation from the second equation givesA=-1=B=-2.Therefore, we have...Z3x2+ 6x+ 2x2+ 3x+ 2dx=Z3 +-3x-4x2+ 3x+ 2dx=Z3 +-1x+ 1+-2x+ 2dx= 3x-ln|x+ 1| -2 ln|x+ 2|+c(b)Zln (x+ 1)x2dxNeed to use integration by parts...u= ln (x+ 1)dv=x-2dxdu=1x+ 1dxv=-1xThis gives...Zln (x+ 1)x2dx=-ln (x+ 1)x-Z-1x1x+ 1dx=-ln (x+ 1)x+Z1x(x+ 1)dx
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Math 1620Exam 2 Practice Problems SolutionsFall 2019Now, for the second integral, we need to use partial fraction decomposition...1x(x+ 1)=Ax+Bx+ 1=A(x+ 1) +B(x)x(x+ 1)=1 =A(x+ 1) +B(x) =x(A+B) + (A)=A+B= 0 andA= 1=B=-1Now, we have...Zln (x+ 1)x2dx=-ln (x+ 1)x-Z-1x1x+ 1dx=-ln (x+ 1)x+Z1x(x+ 1)dx=-ln (x+ 1)x+Z1x-1x+ 1dx=-ln (x+ 1)x+ ln|x| -ln|x+ 1|+c(c)Zsin3(θ) cos4(θ)Zsin3(θ) cos4(θ)=Zsin2(θ) sin (θ) cos4(θ)=Z(1-cos2(θ)) sin (θ) cos4(θ)Letu= cos (θ), thendu=-sin (θ).=Z-(1-u2)u4du=Z(-u4+u6)du=-15u5+17u7+c=-15cos5(x) +17cos7(x) +c
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Math 1620Exam 2 Practice Problems SolutionsFall 2019(d)Ztt4+ 2dtLetu=t2. Thendu= 2tdt. Now we have...Ztt4+ 2dt=Z121u2+ 2du=122tan-1u2+c=122tan-1t22+c(e)Zdx4x2-1dxNeed to use a trig.substitution...on the right triangle, let 2xbe the hypotenuse,4x2-1 be the ”opposite” side, and 1 be the ”adjacent” side.Then sec (θ) = 2x=x=12sec (θ), tan (θ) =4x2-1, anddx=12sec (θ) tan (θ).Then the integral becomes...Zdx4x2-1dx=Z1tan (θ)12sec (θ) tan (θ)=12Zsec (θ)=12ln|sec (θ) + tan (θ)|+c=12ln|2x+4x2-1|+c
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Math 1620Exam 2 Practice Problems SolutionsFall 2019(f)Z311 +x2x2dxNeed to use a trig. substitution...on the right triangle, let1 +x2be the hypotenuse,xbe the ”opposite” side, and 1 be the ”adjacent side.Then tan (θ) =x,dx=sec2(θ), sec (θ) =1 +x2,θ= tan-1(x),θ(1) =π4, andθ(3) =π3.Now, wehave...Z311 +x2x2dx=Zπ/3π/4sec (θ)tan2(θ)sec2(θ)=Zπ/3π/41/cos (θ)sin2(θ)/cos2(θ)sec2(θ)=Zπ/3π/4cos (θ)sin2(θ)(1 + tan2(θ))=Zπ/3π/4csc (θ) cot (θ)+Zπ/3π/4cos (θ)sin2(θ)tan2(θ)=-csc (θ)π/3π/4+Zπ/3π/4cos (θ)sin2(θ)sin2(θ)cos2(θ)=-23+12/2+Zπ/3π/4sec (θ)=2-23+ ln|sec (θ) + tan (θ)|π/3π/4=2-23+ ln|2 +3| -ln|2 + 1|(g)Zx2x2+ 7x-4dxNeed to use partial fraction decomposition...x2x2+ 7x-4=x(2x-1)(x+ 4)=A2x-1+Bx+ 4=A(x+ 4) +B(2x-1)(2x-1)(x+ 4)=x=A(x+ 4) +B(2x-1) =x(A+ 2B) + (4A-B)=A+ 2B= 1 and 4A-B= 0Subtracting 2 times the second equation to the first equation gives 9A= 1 =A=19
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Math 1620Exam 2 Practice Problems SolutionsFall 2019andB=49. Now, we have...Zx2x2+ 7x-4dx=Z1/92x-1+4/9x+ 4dx=118ln|2x-1|+49ln|x+ 4|+c(h)Ztan2(θ) sec4(θ)Ztan2(θ) sec4(θ)=Ztan2(θ) sec2(θ) sec2(θ)=Ztan2(θ)(1 + tan2(θ)) sec2(θ)

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