Consider the area between the graphs x+3y=32 and x+8=y2. This area can be computed in two different ways using integrals
First of all it can be computed as a sum of two integrals
where a= , b=, c= and
f(x)=
g(x)=
I found a, b, c. But I can't seem to figure out f(x) and g(x). I would assume f(x) is (32-x)/3 and g(x) is (x+8)^(1/2), but it's not right.. Can you help me?
First of all it can be computed as a sum of two integrals
where a= , b=, c= and
f(x)=
g(x)=
I found a, b, c. But I can't seem to figure out f(x) and g(x). I would assume f(x) is (32-x)/3 and g(x) is (x+8)^(1/2), but it's not right.. Can you help me?
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These types of problems you should draw the graph out.
Never assume what area the problem is referring to until you've drawn the graphs and considered any region(s) bounded by the graphs. Sometimes there may even be more than 1 region, such as areas bounded by 3 curves or more, or considering trig functions.
The fact that you have y^2 in the second equation makes y no longer a function
because y = + or - sqrt(x+8).... two outputs for every input of x in the domain.
Here's what the graph looks like:
http://www.wolframalpha.com/input/?i=gra…
It would be a lot simpler to do f(y) instead of f(x), because then you would not to break the region into different parts. But for f(x), it looks like the first region has the integral equation:
Integral from a to b { sqrt(x+8) - [-sqrt(x+8)] } dx
or simply
Integral from a to b 2sqrt(x+8)dx
while the second region has the area equation:
Integral from b to c { (32-x)/3 - [-sqrt(x+8)] } dx
You then add these integrals together.
Here, a is the x-coordinate of the vertex of the sideways parabola, b is that of the left point of interception between the graphs, and c is the right point of interception.
If you were to consider functions in terms of y, then you will get:
Integral from a to b [ 32-3y - (y^2 - 8) ] dy
Here, a is y-coordinate of the bottom interception. b is that of the top interception.
The integral is a lot simpler (no need to add two integrals), but may feel strange because now x is the function and y is the variable. (You'll get plenty more practice on stuff like this in multivariable calculus).
Never assume what area the problem is referring to until you've drawn the graphs and considered any region(s) bounded by the graphs. Sometimes there may even be more than 1 region, such as areas bounded by 3 curves or more, or considering trig functions.
The fact that you have y^2 in the second equation makes y no longer a function
because y = + or - sqrt(x+8).... two outputs for every input of x in the domain.
Here's what the graph looks like:
http://www.wolframalpha.com/input/?i=gra…
It would be a lot simpler to do f(y) instead of f(x), because then you would not to break the region into different parts. But for f(x), it looks like the first region has the integral equation:
Integral from a to b { sqrt(x+8) - [-sqrt(x+8)] } dx
or simply
Integral from a to b 2sqrt(x+8)dx
while the second region has the area equation:
Integral from b to c { (32-x)/3 - [-sqrt(x+8)] } dx
You then add these integrals together.
Here, a is the x-coordinate of the vertex of the sideways parabola, b is that of the left point of interception between the graphs, and c is the right point of interception.
If you were to consider functions in terms of y, then you will get:
Integral from a to b [ 32-3y - (y^2 - 8) ] dy
Here, a is y-coordinate of the bottom interception. b is that of the top interception.
The integral is a lot simpler (no need to add two integrals), but may feel strange because now x is the function and y is the variable. (You'll get plenty more practice on stuff like this in multivariable calculus).