Calculate the area between 2 curves integration 4 questions 

 1/ between 3 and 1   (x+1)^2 d.x

 2/ between 4 and 1   (2x+1)(x-2) d.x

 3/ between + 0.001 and -0.001   50 cos(50 pi t) d.t

 4/ between pi/3 and 0   10 sin (2t- pi/3)d.t   

                                thanks, please show all working

Ninja1 –

Here goes. . .

Problem #1
=========

f(x) = (x+1)^2  dx = (x^2 + 2x + 1) dx

The integral of that function is:
1/3*x^3 + x^2 + x + c (where c is a constant) ; evaluated on the range
between 1 and 3, the function is always positive so --
1/3 *1 + 1 + 1 = 2.333
1/3 3^3 + 3^2 +3 = 9 + 9 + 3 = 12

Area between the two points = 12 – 2.33 = 9.67


Problem #2
==========

f(x) = (2x+1) * (x-2) dx = (2x^2 –3x – 2) dx

The integral of that function is:
2/3*x^3 – 3/2*x^2 –2x + c; evaluated between 1 and 4; this function
becomes negative between x=2 and x = -1/2, so we must evaluate two
ranges –
from 1 to 2 and from 2 to 4:

From 1 to 2:
2/3* 1 – 3/2*1 – 2 = -2.83 (a negative area because it's below the
x-axis)
2/3*8 – 3/2*4 – 4 = -4.67 
In the negative range, total area = -4.67 + 2.83 = -1.84

From 2 to 4:
We've already calculated 2: -4.67
2/3*64 – 3/2*16 – 2*4 = 42.67 – 24 – 8 = 10.67
In the positive range, the area between the two points = 10.67 + 4.67
= 15.34

TOTAL AREA =  15.34 + 1.84 = 17.18


Problem #3
=========

The integral of cos(t) dt = sin(t) dt + c

For f(x) = 50 cos(pi*t) dt , the integral is:
50 sin(pi*t) + c

Thus the values, evaluated between -0.001 and 0.001 will be:
50 sin (-0.0031) = -0.0027 
50 sin (0.0031) = -0.0027
The sum of the two symmetric areas is 0.0027 and ABS VAL (-0.0027) =
0.0054


Problem #4

The integral of sin(t) dt = - cos(t) + c

For f(x) = 10 sin (2t – pi/3) dt, the integral is:
-10 cos (2t-pi/3) + c

This function goes crosses the x-axis at t = pi/6, which is in the
middle of the range we're measuring.  First we'll evaluate the range
pi/6 to pi/3:
-10 cos (pi/3 – pi/3) = -10 cos (0) = -10
-10 cos (2/3 pi – pi/3) = -10 cos(pi/3) = -10 cos (1.047) = -10
(.9998) = - 10 (to 2 decimal places)

Area from pi/6 to p/3 is 0

Now from 0 to pi/6:
-10 cos (-pi/3) = -10
The value at pi/6 = -10

Area from 0 to pi/6 is zero.

Total area zero.  You might ask why?  It's because the cos (0) and cos
(1.047) are so close that calculating them to three decimal places
yields the same number.  The cosine function is virtually flat in this
range, producing no appreciable value under the curve.

If there are any questions about these answers please let me know
before rating this answer.  Working with powers and mathematical
notation in Google Answers can sometimes be confusing.

Best regards,

Omnivorous-GA

---

Clarification of Answer by omnivorous-ga on 07 Jan 2003 03:45 PST

Ninja --

Please note that the addition below for problem #1 should be:
1/3 3^3 + 3^2 + 3 = 9 + 9 + 3 = 21

Area between the two points = 21 – 2.33 = 18.67

---

Clarification of Answer by omnivorous-ga on 07 Jan 2003 11:33 PST

The integrals for #3 and #4 given yesterday failed to properly use the
integration formulas.  Corrected integrals are:

Problem #3
=========

The integral of cos(at) dt = 1/a sin(at) dt + c

For f(x) = 50 cos(pi*t) dt , the integral is:
50/pi sin (pi * t) + c

Thus the values, evaluated between -0.001 and 0.001 will be:
15.92 * sin (-0.0031) = -0.0009
15.92 * sin (0.0031)  =  0.0009

Total area = 0.0017


Problem #4
=========

The integral of sin (t) dt = - cos(t) + c

For f(x) = 10 sin (2t – pi/3) dt, the integral is has to be calculated
with the chain rule:
10 sin (u) du, where u = (2t-pi/3); du = 2 dt; ½ du = dt

The integral becomes: 
-½ cos (2t – pi/3) + c 

Thus the values on the range pi/6 to pi/3 are:
-1/2 cos (0) = -0.5000
-1/2 cost (pi/3) = .-0.499

Area of the range = .0001

The same area applies between pi/6 and 0  - -

Total area = 0.0002

Best regards,

Omnivorous-GA

---

Clarification of Answer by omnivorous-ga on 07 Jan 2003 17:19 PST

My daughter tells me that I've made a mistake in evaluating the sin by
using degrees instead of RADIANS.  So, corrected calculations are:
 
Problem #3 
========= 
 
The integral of cos(at) dt = 1/a sin(at) dt + c 
 
For f(x) = 50 cos(pi*t) dt , the integral is: 
50/pi sin (pi * t) + c 
 
Thus the values, evaluated between -0.001 and 0.001 will be: 
15.92 * sin (-0.0031) = -0.1571 
15.92 * sin (0.0031)  =  0.1571 
 
Total area = 0.3142
 
 
Problem #4 
========= 
 
Similarly with problem 4 – it needs be adjusted for RADIANS and a math
error in my final forumula:

The integral of sin (t) dt = - cos(t) + c 
 
For f(x) = 10 sin (2t – pi/3) dt, the integral is has to be calculated
with the chain rule:
10 sin (u) du, where u = (2t-pi/3); du = 2 dt; ½ du = dt 
 
The integral becomes:  
- 5 cos (2t – pi/3) + c  
 
Thus the values on the range pi/6 to pi/3 are: 
- 5 cos (0) = -5.000 
- 5 cos (pi/3) = -2.5 
 
Area of the range = 2.5 
 
The same area applies between pi/6 and 0  - - 
 
Total area = 5.0 
 
Thanks to researcher rbnn-ga for keeping this answer accurate.

---

Clarification of Answer by omnivorous-ga on 08 Jan 2003 05:36 PST

Ninja01 --

Note that Rbnn is correct about the factor missing in #3:

Problem #3  
=========  
  
The integral of cos(at) dt = 1/a sin(at) dt + c  
  
For f(x) = 50 cos(50pi*t) dt , the integral is:  
1/pi sin (pi * t) + c  
  
Thus the values, evaluated between -0.001 and 0.001 will be:  
.3185 * sin (-0.0031) = -0.0031  
.3185 * sin (0.0031)  =  0.0031  
  
Total area = 0.0063

---

Clarification of Answer by omnivorous-ga on 08 Jan 2003 12:59 PST

Correction to Problem #3   
=========================
   
The integral of cos(at) dt = 1/a sin(at) dt + c   
   
For f(x) = 50 cos(50pi*t) dt , the integral is:   
1/pi sin(50pi * t) + c =
.3185 * sin (157.08t)   
   
Thus the values, evaluated between -0.001 and 0.001 will be:   
.3185 * sin (-0.1571) =-(0.3185 * 0.4738) = -0.1509 
.3185 * sin (0.1571)  = 0.3185 * .4738 = .1509
   
Total area = 0..3018

excellent

The indefinite integral in problem 4 is incorrect. 

The value of cos(pi/3) in problem 4 is incorrect.

I cannot evaluate the accuracy of the solutions as a whole because I
am not sure what questioner is asking in his phrase "the area between
2 curves" - I don't know what two curves he's talking about.

Also, both the indefinite integral and the numerical computations in
problem 3 are incorrect (as they were in problem 4).

I don't understand the answer in problem 2, but I would be surprised
if it is correct - it could be.

The computations in problem 1 are also wrong, as the value of the
indefinite integral at 3 is wrong.

I would try to give the correct answer here but I don't understand the
question. I do understand however that the reasoning given in both
questions is incorrect.

I agree with the analysis in the clarification (7 Jan 3:45 PST) for
problem 1, but my analysis of problems 2, 3, and 4 continues to differ
with answers and with clarifications:

In the clarification (7 Jan 11:33 PST)  and the original answer for
problem 3, the question is asking about a different function than that
for which the computation in the answer is being performed.

In that same clarification for problem 3, the values used for
sin(-.0031) and  for sin(-.0031) are each incorrect.

The indefinite integral in the clarification (7 Jan 11:33 PST) in
problem 4 is still incorrect. This can be verified by differentiation.

The value used for cos(pi/3) in the clarification in problem 4 is
incorrect.

Regarding the clarification of 7 Jan 2003 17:19 PST:  1,2, and 4 seem
ok. Note that calculus on trigonmetric functions is always performed
in radians - otherwise the derivative of sin would not be cosine.

It is still the case, however, that my reading of the question
suggests that the function, 50 cos (50 pi t),  whose area is requested
in part 3 of the question is different from the function, 50 cos (pi
t),  whose integral is computed in part 3 of the answer.

The recent clarification of 8 Jan 2003 5:36 PST states that the
integral of 50 cos(50 pi t) is 1/pi sin (pi t) .

However, the indefinite integral of 50 cos (50 pi t) is  1/pi sin (50
pi t). This may be verified by differentiation.