# Resolving an Integral by Upper and Lower Sums

So, the problem statement says that i have to determinate the Upper and Lower Sums that aproximate the area under the graph given by the next function: $$f(x) = x^3$$ in the interval[0,1] with a partition of 0,2

So, i preoceeded to determinate the Upper and Lower Sums but I dont come up with the righ answer (i know because i corroborated by getting the resault of the integral $\int x^3\, dx$ betwen 0 and 1, with my calculator)

P={0; 0,2; 0,4; 0,6; 0,8;1}
$L(f,P) = \sum^{5}_{i=0}$mi(ti-ti-1) = $(0^3)(0,2) + (0,2^3)(0,2) + (0,4^3)(0,2) + (0,6^3)(0,2) + (0,8^3)(0,2) = 0,10$

That is just plain wrong but i dont know what im doing wrong...well i wont redact how i did the Upper sums because i guess you got the point...

Thanks.

Last edited:

SammyS
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So, the problem statement says that i have to determinate the Upper and Lower Sums that approximate the area under the graph given by the next function: $$f(x) = x^3$$ in the interval[0,1] with a partition of 0,2

So, i proceeded to determinate the Upper and Lower Sums but I don't come up with the righ answer (i know because i corroborated by getting the result of the integral $\int x^3\, dx$ between 0 and 1, with my calculator)

P={0; 0,2; 0,4; 0,6; 0,8;1}
$L(f,P) = \sum^{5}_{i=0}$mi(ti-ti-1) = $(0^3)(0,2) + (0,2^3)(0,2) + (0,4^3)(0,2) + (0,6^3)(0,2) + (0,8^3)(0,2) = 0,10$

That is just plain wrong but i dont know what im doing wrong...well i wont redact how i did the Upper sums because i guess you got the point...

Thanks.
I get 0.016 --- or as you write it, 0,016 .

I get 0.016 --- or as you write it, 0,016 .

Yes, sorry I posted the answer wrong, its 0.16, but what concerns me is that if i did it ok or is wrong the preocedure...

SammyS
Staff Emeritus