Saturday, July 25, 2015

`int_-1^1(|3^x - 2^x|)dx` Evaluate the integral and interpret it as the area of a region. Sketch the region.

Since `3^x>2^x` for `x>0` and `3^x<2^x` for `x<0` we have


`int_-1^1|3^x-2^x|dx=int_-1^0(2^x-3^x)dx+int_0^1(3^x-2^x)dx=`  


Now we apply formula `int a^x=a^x/ln a+C.`  


`(2^x/ln 2-3^x/ln 3)|_-1^0+(3^x/ln3-2^x/ln2)|_0^1=`


`1/ln2-1/ln3-1/(2ln2)+1/(3ln3)+3/ln3-2/ln2-1/ln3+1/ln2=`  


`(-3ln3+8ln2)/(6ln2ln3)=`


`ln(256/27)/(ln3ln64)approx0.4923`                                                                                    

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