[h=1]Use the limiting process to find the area between the graph of the function?[/h]Use the limiting process to find the area between the graph of the function f(x)=x^2+x^3 and the x-axis over the interval [−2,0].
Area =
THis is how I worked it out below but its still not right..what am I doing wrong
function is negative on [-2,-1] and positive on [-1,0]
set s1 the are between the curve, x axis and lines x = -1, x = 0
and s2 the area between the curve, x axis and lines x=-2,x=-1
Requested area is
Area = s1 - s2
divide interval [-1,0] in n intervals width = 1/n
and let's compute the height of rectangles
which is the value of the function at
x = - k/n for k=0..n
f(-k/n) = k^2/n^2 -k^3/n^3
the area of the generic rectangle is
(k^2/n^2 -k^3/n^3)(1/n) = k^2/n^3 - k^3/n^4
and
s1 = lim Σ[k^2/n^3 - k^3/n^4] (as n--> infinity)
Σk^2/n^3 = (1/n^3) Σk^2 = (n + 1)(2n + 1)/(6n^2)
Σk^3/n^4 = (1/n^4)Σk^3 = (n + 1)^2/(4n^2)
s1 = lim [(n + 1)(2n + 1)/(6n^2) - (n + 1)^2/(4n^2)] =
= lim [(n + 1)(n - 1)/(12n^2) = 1/12 as n --> infinity
in the same way you find
s2 = -17/12
finally area = 1/12 - (-17/12) = 18/12 = 3/2
Area =
THis is how I worked it out below but its still not right..what am I doing wrong
function is negative on [-2,-1] and positive on [-1,0]
set s1 the are between the curve, x axis and lines x = -1, x = 0
and s2 the area between the curve, x axis and lines x=-2,x=-1
Requested area is
Area = s1 - s2
divide interval [-1,0] in n intervals width = 1/n
and let's compute the height of rectangles
which is the value of the function at
x = - k/n for k=0..n
f(-k/n) = k^2/n^2 -k^3/n^3
the area of the generic rectangle is
(k^2/n^2 -k^3/n^3)(1/n) = k^2/n^3 - k^3/n^4
and
s1 = lim Σ[k^2/n^3 - k^3/n^4] (as n--> infinity)
Σk^2/n^3 = (1/n^3) Σk^2 = (n + 1)(2n + 1)/(6n^2)
Σk^3/n^4 = (1/n^4)Σk^3 = (n + 1)^2/(4n^2)
s1 = lim [(n + 1)(2n + 1)/(6n^2) - (n + 1)^2/(4n^2)] =
= lim [(n + 1)(n - 1)/(12n^2) = 1/12 as n --> infinity
in the same way you find
s2 = -17/12
finally area = 1/12 - (-17/12) = 18/12 = 3/2