No, they intersect at only a single point, and it's not at either of those values of x.Chandasouk said:Yeah.
Let f(x) = 20 + X + X2
Let g(x) = X2-5X
Find the area of the region enclosed by the two graphs.
The graphs intersect at X = -2 and 5, so those are the bounds.
Chandasouk said:F(x) is the upper curve.
So I take the definite integral of
[(20 + X - X^2) - (X^2-5X)dx] from [-2, 5]
The answer is 343/3, but I do not get that answer when I compute the integral at all
The expression posted in the thread title contains an x3 term, so this isn't even the same problem.Chandasouk said:Yeah.
Let f(x) = 20 + X + X2
Let g(x) = X2-5X
Find the area of the region enclosed by the two graphs.
No, no, no! The two [itex]x^2[/itex] terms cancel they don't add!Chandasouk said:The problem in my Calc book just stated
Let f(x) = 20 + X + X^2
Let g(x) = X^2-5X
Find the area of the region enclosed by the two graphs.
I set these two equal to each other
20 + X + X^2 = X^2-5X
Through Algebra, I found that
0 = 2X^2 - 6X -20
which factors to
2(X-5)(X+2)
Thus the curves intersect at X = -2 and X = 5. I made a mistake earlier saying that the graphs intersected at those two points; sorry.
f(x) is the upper curve. From there they just used used the formula for calculating the area between curves and obtained the answer 343/3 but i cannot work out the math to get that
Chandasouk said:Homework Statement
I get the answer 19/3 but the real answer is 343/3. Can you show work to how it is that? I did this many times and still came up with 19/3 instead
I got up to evaluating the integral part of finding the area between two curves but can never get the correct answer. I have a feeling I'm messing up signs somewhere.
Looks like you have set the integral up correctly. Can you show the rest of your work? I am getting 343/3 as well.Chandasouk said:f(x) is the upper curve. From there they just used used the formula for calculating the area between curves and obtained the answer 343/3 but i cannot work out the math to get that
A definite integral is a mathematical concept used to calculate the area under a curve or the accumulation of a function over a given interval. It is represented by the symbol ∫ and has both an upper and lower limit.
Evaluating a definite integral allows us to find the exact numerical value of the area under a curve or the accumulation of a function, which can be useful in many real-world applications such as calculating distances, volumes, and probabilities.
To evaluate a definite integral, we use the fundamental theorem of calculus, which states that the definite integral of a function f(x) from a to b is equal to the difference of the antiderivative of f(x) at b and a, or ∫ab f(x) dx = F(b) - F(a).
The function in this definite integral is 20x + 3x^2 - (2/3)x^3.
To solve this definite integral, we first find the antiderivative of the function, which is 10x^2 + x^3 - (1/9)x^4. Then, we substitute the upper and lower limits of -2 and 5 into the antiderivative and find the difference, resulting in a final value of 375.111.