Question 1179691
<br>
This problem can be solved easily without calculus.  The given function is a straight line, so the area under the curve on the interval [1,b] is a trapezoid.  The area of a trapezoid is<br>
{{{((a+b)/2)(h)}}}<br>
where h is the height and a and b are the lengths of the two bases.<br>
For this problem, the height is (b-1); the two bases are f(1)=12 and f(b)=5b+7.<br>
{{{((a+b)/2)(h)=((5b+19)/2)(b-1)=88}}}
{{{(5b+19)(b-1)=176}}}
{{{5b^2+14b-19=176}}}
{{{5b^2+14b-195=0}}}<br>
Note this is the same equation the other tutor ends up with using calculus.<br>
She then goes on, as she always does, to magically separate the middle term 14b into two parts, producing a quadratic with four terms that can be factored by grouping.  That's fine for showing the solution -- but it does nothing to teach the student how the factoring is done.<br>
So let me take a bit of time here to talk about ways to factor a quadratic like this.  There are many methods that are taught; most of them that I have seen consist of well defined steps that mysteriously lead to the right factorization.<br>
I prefer one that calls on the student to use his powers of reasoning to obtain the answer.<br>
In this case we want a factorization of the form<br>
{{{5b^2+14b-195=(pb+q)(rb+s)}}}<br>
Since the b^2 term comes from (pb)(rb), and the coefficient of b^2 in the quadratic is 5, clearly p and r have to be 5 and 1, in some order.  We can assume p=5 and r=1, giving us<br>
{{{5b^2+14b-195=(5b+q)(1b+s)}}}<br>
Now look at the constant term in the quadratic: -195.  That constant comes from q times s: {{{qs=-195}}}<br>
We can see that q and s are of opposite sign; and we can see that one of them contains 5 as a factor.<br>
Here is the where the logical reasoning makes what looks like a big problem much easier.  If q contained a factor of 5, then the two terms in the binomial (pb+q) would have a common factor of 5.  That would make the quadratic have a common factor of 5; but it does not.  So q cannot contain a factor of 5; so s must contain a factor of 5.<br>
-195 is 5(-39); at this point we can play with the signs of the two binomial factors and the possible values of q and s to find the combination that gives us the correct middle term.<br>
With a little playing around like that, we find<br>
{{{5b^2+14b-195=(5b+39)(1b-5)}}}<br>
That gives us possible values of -39/5 or 5 for b; the statement of the problem asking us to find the area under the curve on the interval [1,b] means we choose the positive value for b.<br>
So b=5, and now we have all we need to find the area as the area of a trapezoid; its height is b-1=4; the bases are f(1)=12 and f(5)=32; the area is<br>
{{{((a+b)/2)(h) = ((12+32)/2)(4) = 22*4 = 88}}}<br>