How do you use limits to evaluate #int 8xdx# from [0,2]?

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Answer 1 · 2016-12-06T21:41:41

Evaluating an integral using limits will give you the formula:

#int_a^bf(x)dx=lim_(n->oo)sum_(i=1)^nf(x_i)*Deltax#

Where:
#Deltax=(b-a)/n#

#x_i=a+iDeltax#

So for:

#int_(0)^2 8xdx#

#Deltax=2/n#

#x_i=(2i)/n#

#=lim_(n->oo)sum_(i=1)^n8((2i)/n)*2/n#

#=lim_(n->oo)sum_(i=1)^n(32i)/n^2#

We can pull out the constant:

#=lim_(n->oo)32/n^2sum_(i=1)^ni#

#i=(n(n+1))/2#

#=lim_(n->oo)32/n^2*(n(n+1))/2#

Now, here you can skip some tedious algebra:

Since the degree of the denominator is 2 and the degree of the numerator is also two, the limit will just be a ratio:

Thus, #=lim_(n->oo)32/n^2*(n(n+1))/2=16#