# lecture8_1.pdf - 8.1 Integration by Parts Example Evaluate...

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8.1 Integration by Parts
Example: EvaluateZxcosxdx
Example: EvaluateZxcosxdxNote thatZxcosxdx=Zx(sinx)0dxIt seems not that easy to solve it. But,Z(x)0sinxdxis easy to evaluate.
Example: EvaluateZxcosxdxNote thatZxcosxdx=Zx(sinx)0dxIt seems not that easy to solve it. But,Z(x)0sinxdxis easy to evaluate.Is there any relation betweenRx(sinx)0dxandR(x)0sinxdx?
Example: EvaluateZxcosxdxNote thatZxcosxdx=Zx(sinx)0dxIt seems not that easy to solve it. But,Z(x)0sinxdxis easy to evaluate.Is there any relation betweenRx(sinx)0dxandR(x)0sinxdx?Indeed, by the product rule,(xsinx)0=x0sinx+x(sinx)0
Therefore,Zxcosxdx=Zx(sinx)0dx=Z(xsinx)0-x0(sinx)dx=xsinx-Zsinxdx=xsinx+ cosx+C.The technique used here is called “integration by parts”.In general, the product rule says,ddx(f(x)g(x)) =f0(x)g(x) +f(x)g0(x).
Thus,Zddx(f(x)g(x))dx=Zf0(x)g(x)dx+Zf(x)g0(x)dxf(x)g(x) =Zf0(x)g(x)dx+Zf(x)g0(x)dxTherefore,Zf(x)g0(x)dx=f(x)g(x)-Zf0(x)g(x)dx.This is called the integration-by-parts formula.Letu=f(x)andv=g(x), thendu=f0(x)dxanddv=g0(x)dx.Therefore,Zudv=uv-Zvdu
Example: FindRxlnxdx.
Example: FindRxlnxdx.First try:u=x, dv= lnxdxdu=dx, v=?Not so clear here.
Example: FindRxlnxdx.First try:u=x, dv= lnxdxdu=dx, v=?Not so clear here.Second try:u= lnx, dv=xdxdu=dxxandv=x22Zxlnxdx=x22lnx-Zx22·dxx=x22lnx-Zx2dx=x22lnx-x24+C.
Example: FindRlnxdx
Example: FindRxlnxdx.First try:u=x, dv= lnxdxdu=dx, v=?Not so clear here.Second try:u= lnx, dv=xdxdu=dxxandv=x22Zxlnxdx=x22lnx-Zx22·dxx=x22lnx-Zx2dx=x22lnx-x24+C.
Example: FindRlnxdx
u= lnx, dv=dxdu=dxx, v=x.So,Zlnxdx=xlnx-Zx·1xdx=xlnx-x+C.
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