1)
a) y =
2t + 1
ð
Y/1 + t/(-1/2) = 1
Which gives y intercept = 1 and t intercept = -½, hence the given line
will cut the y axis at (0, 1) and the t axis at (-1/2, 0). Making use of these
intercepts the given line is drawn below:
The area under this line above t-axis and between the vertical lines t=1
and t=3 has been shaded as below:
The area of shaded region can be evaluated geometrically, which is,
Area(ABCDE) = area of triangle ABE + area of rectangle BCDE
= ½ * AE *
BE + CD * CB
= ½ * (AD –
ED) * CD + CD*CB
= ½ * (7 –
3) * (3-1) + (3-1)*3
= ½ * 4*2
+ 2*3
= 10 square
units
Hence, the required area is 10 square units.
1)
b) The area between t = 1 and t = x is sketched
as below:
In the above sketch, the point D on the t-axis has been taken as (x,0).
Hence, point A = (x, 2x+1). This gives AD = 2x + 1.
Also, point C = (1,0) and hence point B = (1,3). Hence, BC = 3-0 = 3 =
ED.
This gives AE = AD – ED = (2x+1) – 3 = 2x - 2
The area of the required part = A(x) = area (ABCDE)
= area of triangle ABE + area of rectangle BCDE
= ½ * BE * AE + CD*ED
= ½ * CD * AE + CD*BC
(Since BE = CD and ED = BC)
= ½*(x-1)*(2x-2) + (x-1)*3
= (x-1)*[1/2(2x-2) + 3]
= (x-1)*[x-1+3]
= (x-1)*(x+2)
= x^2 +x – 2
Hence, A(x) = x^2 + x - 2.
1)
c)
using the result of part B we have
A(x) = x^2 + x -2
On differentiating A(x) w.r.t. x, we get
A’(x) = d(x^2+x-2)/dx = d(x^2)/dx + d(x)/dx – d(2)/dx
= 2x + 1 – 0 = 2x+1
That means the derivative of the area function A(x) is 2x+1, which is the
same expression as the given line (y = 2t+1) below which we had to find the
area. Hence, the differentiation of area function gives back the original line.
2) a)
The given region
A(x) is the area enclosed between the function y = t^2+1 and t axis from t = -1
to t = x as shown below:
2 b) A(x) = ʃx-1(1+t^2)dt = ʃx-1
dt + ʃx-1 t^2dt
= x-(-1) + [x^3 – (-1)^3]/3
= x+1 +
(x^3+1)/3
= (x^3
+ 3x + 4)/3
2 c) A(x) = (x^3 +3x+4)/3
Hence, A’(x) = d[(x^3
+3x+4)/3]/dx
=
1/3[d(x^3)/dx + d(3x)/dx + d(4)/dx]
= 1/3[3x^2
+ 3 + 0]
=
x^2 + 1
We get A’(x) = x^2 + 1 i.e. A’(t) = 1+t^2 which is the same
expression as the integrand of part A.
2 d) A(x+h) is the region between y = 1+t^2 and t-axis from t = -1
to t = x+h while A(x) is the region between y = 1+t^2 and t –axis from t = -1
to t = x. Hence, A(x+h) – A(x) will effectively be the region between = 1+t^2
and t-axis from t = x to t = x+h as shown below in the sketch.
2 e) Below is the rectangle that approximate A(x+h) – A(x)
for very small h.
Now,
From part B, A(x) = (x^3 +3x+4)/3
So, A(x+h) = [(x+h)^3 +3(x+h)+4]/3
Hence, A(x+h) – A(x) = 1/3
[(x+h)^3 – x^3 + 3(x+h) -3x + 4 -4]
=
1/3[x^3 + h^3 +3x^2h + 3xh^2 –x^3 +3h]
= 1/3[h^3
+ 3(hx^2 +xh^2 +h)]
= h[h^2/3
+ (x^2 +xh + 1)]
So, [A(x+h) – A(x)]/h = h^2/3 + (x^2 +xh + 1)
But as h is very small so taking h = 0 we get
[A(x+h) – A(x)]/h = h^2/3 + (x^2 +xh + 1) = 0 + (x^2 + 0 + 1) =
x^2 + 1
Hence, the expression.
2 f) As we saw in part E that rate of change of A(x) i.e. [A(x+h)
– A(x)]/h when h is really small becomes x^2 + 1. This explains why A’(x) came
equal to x^2 + 1.
3 a) 
3 b) g(x) will start
to decrease when f(x) takes negative value. In the graph of previous part at
approximately x = 1.25, f(x) = 0.
Hence, at x = 1.25, g(x) will start decreasing.
3 c)
x
|
0.2
|
0.4
|
0.6
|
0.8
|
1.0
|
1.2
|
1.4
|
1.6
|
1.8
|
2.0
|
g(x)
|
0.19997
|
0.39898
|
0.59227
|
0.76784
|
0.90452
|
0.97394
|
0.94978
|
0.82552
|
0.63536
|
0.46146
|
3 d) Below are the values of g’(x) in the intervals
calculated by finding slopes of the above graph.
X
|
g’(x)
|
0 – 0.2
|
0.99985
|
0.2 – 0.4
|
0.99505
|
0.4 – 0.6
|
0.96645
|
0.6 – 0.8
|
0.87785
|
0.8 – 1.0
|
0.6834
|
1.0 – 1.2
|
0.3471
|
1.2 – 1.4
|
-0.1208
|
1.4 – 1.6
|
-0.6213
|
1.6 – 1.8
|
-0.9508
|
1.8 – 2.0
|
-0.8695
|
On plotting these we get
This graph of g’ is similar in shape (although its step
function) to that of f(x). In fact smaller the interval of x we take to calculate
g(x) the closer graph of g’(x) resembles that of f(x).
4)
In problem 2, we had lower limit -1 ( not zero) and the
upper limit x (a variable) and we had got A’(x) = 1 + x^2 ( i.e. same as
integrand).
In problem 3, we had lower limit 0 and the upper limit x and
we got g’(x) = f(x) (i.e. same as integrand of g(x)).
Based on these observations it may be concluded that when
g(x) = ʃxa
f(t)dt
then g’(x) = f(x).
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