HW17MATH10AF18.pdf - HW 17 Solutions October 8 2018 Math...

This preview shows page 1 - 3 out of 7 pages.

HW 17 Solutions; October 8 2018Math 10A with Professor Stankova5.84. Performing the substitutionu=x+ 1, du=dxtells us:Z01(1 +x)1/4dx=Z11u1/4duAs 1/41, we know this integral isdivergent(recallR11xadxis divergent ifa1 andconvergent ifa >1).6. Performing the substitutionu=x2+ 2, du= 2xdxtells us:Z0x(x2+ 2)2dx=12Z21u2duAs 2>1, we know that this integral isconvergent. Now we compute the integral:12Z21u2du= limt→∞12Zt21u2du= limt→∞-12ut2= limt→∞14-12t=148. Notice:limn→∞Z-1-ne-2tdt= limn→∞-12e-2t-1-n= limn→∞12(e2n-e2) =Thus this integral isdivergent, as the limit of proper integrals diverges.10. NoticeZ-∞(y3-3y2)dyis convergent if and only if bothZ0-∞(y3-3y2)dyandZ0(y3-3y2)dyare convergent (this is the definition of convergence for two sided improper integrals). Thusconsider the following limit:limt→∞Zt0(y3-3y2)dy= limt→∞(14y4-y3)t0= limt→∞14t4-t3= limt→∞(14-1t)t4=Notice as the limit of proper integrals is divergent,Z0(y3-3y2)dyis divergent, soZ-∞(y3-3y2)dyisdivergent.12. Notice (using the substitutionu=-x, du=-dx2x):limt→∞Zt1e-xxdx= limt→∞-2Z-t-1eudu= limt→∞-2eu-t-1= limt→∞2(1e-1et) =2eSoZ1e-xxdxisconvergentand equal to2e.
1
14.Z-∞cos(πt)dtis convergent if and only if bothZ0-∞cos(πt)dtandZ0cos(πt)dtare converget.Thus consider:limn→∞Zn0cos(πt)dt= limn→∞sin(πt)πn0= limn→∞sin()πNotice this limit does not exist, soZ0cos(πt)dtis divergent, soZ-∞cos(πt)dtisdivergent.16. Let’s consider the integralZ6-trer/3dr. We can use integration by parts on this integral withf(r) =randg0(r) =er/3, thusf0(r) = 1 andg(r) = 3er/3. Then:Z6-trer/3dr= 3rer/36-t-Z6-t3er/3dr= 3rer/36-t-9er/36-t= 18e2+ 3te-t/3-9e2+ 9e-t/3= 9e2+ (3t+ 9)e-t/3Taking the limit ast→ ∞(using l’Hˆopital’s Rule), we get:limt→∞9e2+ (3t+ 9)e-t/3= 9e2+ limt→∞3t+ 9et/3= 9e2+ limt→∞3(1/3)et/3= 9e2ThusZ6-∞rer/3drisconvergentand equal to9e2.18. Let’s consider the integralZt0x3e-x4dx. Using the substitutionu=-x4,du=-4x3dx, wehave:Zt0x3e-x4dx=-14Z-t40eudu=14Z0-t4eudu=14eu0-t4=14(1-e-t4)Taking the limit ast→ ∞, we get limt→∞14(1-e-t4) =14, soZ0x3e-x4dxis convergent andequal to14. But then notice thatx3e-x4is an odd function, soZ0-tx3e-x4dx+Zt0x3e-x4dx= 0,solimt→∞Z0-tx3e-x4dx=limt→∞-Zt0x3e-x4dx=-14.ThusZ0-∞x3e-x4dxis convergent andequal to-14. ThusZ-∞x3e-x4dxis

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture