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Question Number 48078 by cesar.marval.larez@gmail.com last updated on 19/Nov/18

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18

21)∫e^(2−x) dx  t=2−x   dt=−dx  ∫e^t ×−dt  =(−1)e^t +c  =(−1)e^(2−x) +c  24)∫x^2 e^(x^3 −1) dx  t=x^3 −1    so  dt=3x^2 dx  ∫e^t ×(dt/3)  =(1/3)e^t +c  =(1/3)e^(x^3 −1)  +c  20)∫e^(−0.02x) dx  t=−0.02x   dt=(−0.02)dx  ∫e^t ×(dt/((−0.02)))  =(1/((−0.02)))e^t +c  =(1/((−0.02)))e^(−0.02x) +c

21)e2xdxt=2xdt=dxet×dt=(1)et+c=(1)e2x+c24)x2ex31dxt=x31sodt=3x2dxet×dt3=13et+c=13ex31+c20)e0.02xdxt=0.02xdt=(0.02)dxet×dt(0.02)=1(0.02)et+c=1(0.02)e0.02x+c

Commented by cesar.marval.larez@gmail.com last updated on 19/Nov/18

Sir, thank u

Sir,thanku

Commented by cesar.marval.larez@gmail.com last updated on 19/Nov/18

U r an expert

Uranexpert

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