设f(x)={1+x^2 (x=0) ,求∫(1,3) f(x-2)dx
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设f(x)={1+x^2 (x=0) ,求∫(1,3) f(x-2)dx
![设f(x)={1+x^2 (x=0) ,求∫(1,3) f(x-2)dx](/uploads/image/z/8851879-55-9.jpg?t=%E8%AE%BEf%28x%29%3D%7B1%2Bx%5E2+%28x%3D0%29+%2C%E6%B1%82%E2%88%AB%281%2C3%29+f%28x-2%29dx)
取t=x-2,
∫(1,3) f(x-2)dx=∫(-1,1) f(t)dt
=∫(-1,0) f(t)dt+∫(0,1) f(t)dt
=∫(-1,0)[1+t^2]dt+∫(0,1)[e^t]dt
=e+1/3
∫(1,3) f(x-2)dx=∫(-1,1) f(t)dt
=∫(-1,0) f(t)dt+∫(0,1) f(t)dt
=∫(-1,0)[1+t^2]dt+∫(0,1)[e^t]dt
=e+1/3
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