How do you use limits to find the area between the curve #y=x^4# and the x axis from [0,5]?

Answer 1

#625#

With limits you say... Ok.

Imaging we have some function #f(x)# and we want to find the area underneath it in the interval #[a, b]#. We first start by dividing the area into tiny rectangles like so:

This means the height of each rectangle is equal to the value of the function at that point. Refer to #R_1#. The height of #R_1# is equal to #f(x_0)#, and the area of it is equal to #f(x_0)Delta x# where #Delta x# is just the constant width of each rectangle.

Therefore the area under #[a, b]# is in the diagram above is:
#sum_(n=0)^N R_n=sum_(n=0)^N f(a+nDeltax) Deltax=Deltax sum_(n=0)^N f(a+nDeltax)#
Where #N=(|a-b|)/(Deltax)#

We would also say that for smaller and smaller values of #Delta x#, our approximation of the area will become better and better. Sound familiar? Let's use a limit!

#Area=lim_(Delta x -> 0) sum_(n=0)^N f(a+nDeltax) Deltax#

Now that is some ugly stuff. Let's simplify this a bit and that advantage that the interval you want to find is #[0, 5]# which starts at #0# obviously:

#Area=lim_(Delta x -> 0) sum_(n=0)^N f(nDeltax) Deltax#

Another way of describing #Delta x# is #(|a-b|)/N# (see above equations). So, in this case, it's #(|a-b|)/N#. Since we are applying the limit, there's no need for this to be intuitive and we can go ahead and say: #lim_(N -> oo)#
Finally:
#Area=lim_(N -> oo) sum_(n=1)^N f(5/Nn) 5/N#

Oh, did I mention that #n=0# and #n=1# won't matter, but let's make it #1# for the sake of calculations. And there we have it folks, our very own limit for an area.

To do anything with it, we need to expand stuff. Starting with #f(x)# and then factoring out any things that can be factored:

#lim_(N -> oo) sum_(n=1)^N f(5/Nn) 5/N=lim_(N -> oo) 5/N sum_(n=1)^N (5/N)^4n^4#
#=lim_(N -> oo) 5^5/N^5 sum_(n=1)^N n^4#

And we need to expand out the summation:
#sum_(n=1)^N n^4=(1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n#

#lim_(N -> oo) 5^5/N^5 sum_(n=1)^N n^4#
#=5^5 * 1/5 + 0#
#=625#

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Answer 2

See below.

One of the possible realizations for that integral in terms of Riemann sum is

#lim_(n->oo)sum_(k=0)^(k=n) ((5k)/n)^4(5/n)#

we have

#lim_(n->oo)sum_(k=0)^(k=n) ((5k)/n)^4(5/n)=625#
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Answer 3

#625#

It is better to integrate than to use limits in the problem:

#int_0^5[x^4]dx=x^5/5=5^4=625#
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Answer 4

# int_0^5 \ x^4 \ dx = 625#

By definition of an integral, then

# int_a^b \ f(x) \ dx #
represents the area under the curve #y=f(x)# between #x=a# and #x=b#. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles.

That is

# int_a^b \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)#
Here we have #f(x)=x^4# and we partition the interval #[0,5]# using:
# Delta = {0, 0+1*5/n, 0+2*5/n, ..., 0+n*5/n } # # \ \ \ = {0, 5/n, 2*5/n, ..., 5 } #

And so:

# I = int_0^5 \ x^4 \ dx # # \ \ = lim_(n rarr oo) 5/n sum_(i=1)^n \ f(0+i*5/n)# # \ \ = lim_(n rarr oo) 5/n sum_(i=1)^n \ f((5i)/n)# # \ \ = lim_(n rarr oo) 5/n sum_(i=1)^n \ ((5i)/n)^4# # \ \ = lim_(n rarr oo) 5/n sum_(i=1)^n \ (5/n)^4i^4# # \ \ = lim_(n rarr oo) 5/n * 625/n^4sum_(i=1)^n \ i^4# # \ \ = lim_(n rarr oo) 3125/n^5sum_(i=1)^n \ i^4#

Using the standard summation formula:

# sum_(r=1)^n r^4 = 1/30(6n^5+15n^4+10n^3-n) #

we have:

# I = lim_(n rarr oo) 3125/n^5 1/30 (6n^5+15n^4+10n^3-n)# # \ \ = lim_(n rarr oo) 625/6 (6+15/n+10/n^2-1/n^4)# # \ \ = 625/6 lim_(n rarr oo) (6+15/n+10/n^2-1/n^4)# # \ \ = 625/6 (6+0+0-0)# # \ \ = 625#

Using Calculus

If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:

# int_0^5 \ x^4 \ dx = [ x^5/5 ]_0^5 # # " " = 3125/5-0 # # " " = 625#
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Answer 5

To find the area between the curve ( y = x^4 ) and the x-axis from ( x = 0 ) to ( x = 5 ), you can integrate the absolute value of the function over the given interval. The area is given by the integral:

[ \text{Area} = \int_{0}^{5} |x^4| , dx ]

You need to break this integral into two parts because ( x^4 ) changes sign at ( x = 0 ). The integral from ( x = 0 ) to the point where ( x^4 = 0 ) (which is ( x = 0 )) gives the area under the curve from ( x = 0 ) to the x-axis, and the integral from that point to ( x = 5 ) gives the area above the x-axis.

So, you can rewrite the integral as:

[ \text{Area} = \int_{0}^{5} x^4 , dx - \int_{0}^{0} x^4 , dx ]

Evaluate each integral separately:

[ \int_{0}^{5} x^4 , dx = \frac{1}{5}x^5 \bigg|_{0}^{5} = \frac{1}{5}(5)^5 - \frac{1}{5}(0)^5 = \frac{3125}{5} = 625 ]

[ \int_{0}^{0} x^4 , dx = 0 ]

So, the total area is ( 625 ) square units.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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