Problem in attachment

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- Feb 18th 2006, 10:06 AMfrozenflamesCalculus Help
Problem in attachment

- Feb 18th 2006, 10:49 AMticbolQuote:

Originally Posted by**frozenflames**

---your description of the graph does not tally with the figure as shown.

---the area under the graph from x=1 to x=3 is 2.3 sq.units.

---find the area under the graph from x=0 to x=3.

---the F'(x) = f(x) is beyond me.

Okay.

"INT.(1->3)[f(x)]dx" means the area under the graph of f(x) and above the x-axis from x=1 to x=3.

It means also F(3) minus F(1).

And it is equal to 2.3 sq.units as given.

So if we are to find F(3) minus F(0), which is the area under the graph of f(x) and above the x-axis from x=0 to x=3, then we just need to find the area under the graph from x=0 to x=1 and then add that to 2.3 sq.units.

Meaning,

F(3) -F(0) = {F(1) -F(0)} +{F(3) -F(1)}

F(3) -F(0) = {F(1) -F(0)} +2.3

F(3) -F(0) = {2*1} +2.3 -------------F(1) -F(0) is a rectangle that is 2units by 1unit.

F(3) -F(0) = 2 +2.3

F(3) -F(0) = 4.3 sq.units ----------answer.

Or, D from the multiple choice is the answer.

----------------------------------------------------

Ooppss, you are in integration now?

Okay.

F(1) -F(0) = INT.(0->1)[f(x)]dx

Per the graph as shown,

f(x) = 2 -------------(a horizontal line at y=2.)

from x=0 to x=1,

So,

INT.(0->1)[f(x)]dx

= INT.(0->1)[2]dx

= [2x](0->1)

= 2(1) -2(0)

= 2 -0

= 2 sq.units. - Feb 18th 2006, 11:37 AMCaptainBlack
What we need here is the fundamental theorem of calculus, one version

of which is:

If $\displaystyle F'(x)=f(x)$, then:

$\displaystyle

\int_a^b f(x) dx=F(b)-F(a)

$

So:

$\displaystyle

F(3)-F(0)=\int_0^3 f(x) dx

$

Which can be seen from counting squares is greater than 4 and less than 5

so the correct answer is (D) 4.3. Another way of arriving at this answer is

to split the integral into two parts, from 0 to 1 which is clearly 2, and

from 1 to 3 which we are told is 2.3.

RonL