e^x
d/dx [ ∫ ( Int dt] = x- In 2
2
True or false with full explanation
Let us find
ddx[∫2exlnt dt]=∣u=lnt,dv=dt,du=dtt,v=t∣=ddx[tlnt∣2ex−∫2exdt]=ddx[xex−2ln2−ex+2]=xex+ex−ex=xex≠x−ln2.\frac{d}{dx} [ \int_2^{e^x}\ln t\ dt] =|u=\ln t, dv=dt,du=\frac{dt}{t},v=t|= \frac{d}{dx} [t\ln t|_2^{e^x}- \int_2^{e^x} dt]= \frac{d}{dx} [x{e^x}-2\ln 2- e^x+2]=xe^x+e^x-e^x=xe^x\ne x-\ln 2.dxd[∫2exlnt dt]=∣u=lnt,dv=dt,du=tdt,v=t∣=dxd[tlnt∣2ex−∫2exdt]=dxd[xex−2ln2−ex+2]=xex+ex−ex=xex=x−ln2.
Answer: false
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