Problem in attachment
If I understood what you---and the problem #8--- mean, then,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.
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