What is the area bound by the curve
y=(x^4-2)/2 and the x axis from x = 4 to x = 6
i got -3.98 is this correct
not even close. See that 4th power? Those numbers get big!
So, what did you do?
I took the integral of both and subtracted the x(6) from x(4), i redid it and got 50.5 is that it
No Way. Do you know what 6 to the 5th power is? It's 7776.
Why are you reluctant to show your work?
See
https://www.wolframalpha.com/input/?i=integral%5B4..6%5D+(x%5E4-2)%2F2+dx
I don't think you have caught on to the idea of an integral yet...
To find the area bounded by the curve y = (x^4 - 2)/2 and the x-axis from x = 4 to x = 6, you need to use the definite integral.
The formula for finding the area under a curve between two points is:
A = ∫[a, b] f(x) dx
In this case, the curve is y = (x^4 - 2)/2, and the bounds are x = 4 to x = 6.
So, the integral you need to evaluate is:
A = ∫[4, 6] [(x^4 - 2)/2] dx
To calculate this integral, you can use integration techniques such as the power rule.
Integrating (x^4 - 2)/2 using the power rule, you get:
∫ [(x^4 - 2)/2] dx = [1/10 * x^5 - x/2] + C
Now, you can evaluate the integral using the given bounds:
A = [1/10 * 6^5 - 6/2] - [1/10 * 4^5 - 4/2]
= [777.6 - 18] - [64/10 - 2]
= 759.6 - 6.4
= 753.2
So, the area bounded by the curve y = (x^4 - 2)/2 and the x-axis from x = 4 to x = 6 is approximately 753.2 square units.
Therefore, the answer of -3.98 is not correct.