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 = (4x^(1+1) )/(1+1) = (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)= (242)/(2)`

`b^(2)=121`

`b=11`

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