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

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