How do you evaluate the definite integral #int (x^4-7/x^3+5/sqrtx)dx# from #[1,2]#?

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Answer 1 · 2016-10-24T06:47:06

The definite integral is #=-257/40+10sqrt2#

First we calculate the integral, using the formula

#intx^ndx=x^(n+1)/(n+1)+C# #(n!=-1)#

Then the integral #int_1^2(x^4-7/x^3+5/sqrtx)dx#

#=int_1^2(x^4-7x^(-3)+5x^(-1/2))dx#

#=(x^5/5-7x^(-2)/-2+5x^(1/2)/(1/2))#

#=(x^5/5+7/(2x^2)+10sqrtx)_1^2#

#=(32/5+7/8+10sqrt2)-(1/5+7/2+10)#

#=(31/5-21/8-10+10sqrt2)#

#-257/40+10sqrt2#