Question

How would you use the Fundamental Theorem of Calculus to determine the value(s) of b if the area under the graph g(x)=4x between x=1 and x=b is equal to 240?

Answer 1

The Fundamental Theorem of Calculus regarding geometry states that 
$\int\limits^b_a g{(x)} \, dx = F(b)-F(a)$

Where F is the indefinite integral of $g(x)$

The first step is to integrate $g(x)$
$\int\ {4x} \, dx = \frac{4x^{1+1} }{1+1} = \frac{4x^{2} }{2} =2 x^{2}$

Then substitute the value of $b$ and $a=1$ into $2x^{2}$

$[2 (b)^{2}]-[2 (1)^{2}] = 240$
$2b^{2} -2=240$
$2b^{2}=240+2$
$2b^{2}=242$
$b^{2}= \frac{242}{2}$
$b^{2}=121$
$b=11$

Hence the limit of the area under $g(x)$ is between $a=1$ and $b=11$