The AREA A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles

Question

The AREA A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles

A= lim (n->inf) fn = lim (n->inf) [ f(x1)deltax + f(x2)deltax + … + f(xn)deltan)

\vspace1in

a) Use the above definition to determine which of the following expressions represents the area under the graph of f(x)=x^5 from x=0 to x=2

(b) Evaluate the limit that is the correct answer to part (a). You may find the following formula helpful:

1^5 +2^5 + 3^5 + … + n^5 = ∑ n - i=1 i^5 = [(n^2)(n+1)^2(2n^2+2n-1)]/12

(I have no idea what to do. please explain everything. Thank you)

Answer 1

Divide the interval [0,2] into n equal subintervals each of length Δx = (2-0)/n = 2/n

The right endpoints of the subintervals are 2/n, 2(2/n), 3(2/n), ..., n(2/n) = 2

Using the right endpoint of each subinterval, the approximating sum is:

∑_{(k=1 to n)}[f(x_k)Δx] = (2/n)Σ_{(k=1 to n)}[k(2/n)⁵] = (64/n⁶)Σ_{(k=1 to n)}k⁵ = (64/n⁶)[n²(n+1)²(2n²+2n-1)]/12 =

(16/3)[n²(n+1)²(2n²+2n-1) / n⁶]

∫ _{(from 0 to 2)}x⁵dx = (16/3)lim_{(n approaches ∞)}[n²(n+1)²(2n²+2n-1) / n⁶]

The numerator and denominator of the expression enclosed in the brackets have the same degree (6), so the value of the limit is the quotient of the leading coefficients = 2/1 = 2.

So, ∫_{(from 0 to 2)}x⁵dx = (16/3)(2) = 32/3.

Note: A much easier way to do the problem is to use the Fundamental Theorem of Calculus:

∫_{(from 0 to 2)}x⁵dx = (1/6)x⁶ (evaluated from x = 0 to x = 2) = (1/6)[2⁶ - 0⁶] = 64/6 = 32/3

Answer 2

f(x) = x^5 from x = 0 to x = 2

(a) Okay, I'm going to first explain once again the formula in general for you, if you feel like you already know this, read below the dotted line.

So, the first thing that you have to understand is that the formula that you wrote above for area is just saying to add up a bunch of rectangles. When you read: f(x₁)Δx₁ <------this is the area of one rectangle (specifically the first rectangle you drew.) Here f(x₁) represents the height of the rectangle and Δx₁ represents the width of the rectangle. Now to make the math simple, we assume that each rectangle has the same width, that means Δx₁ = Δx₂ = ... = Δx_n, and since each width is the same, you can easily figure out how big the width is with the formula Δx = (b-a)/n where a, and b are the interval you are interested in, so in this case a = 0 and b = 2, and n is the number of rectangles that you are going to create. Think about it for a while, does that formula make sense? If you want to make 10 rectangles and your interval length is 2 (b/c 2 - 0) wouldn't you just divide the length of the interval by 10 making each width 2/10 or 1/5?

Therefore, each of our Δx = (2 - 0)/n = 2/n (don't worry too much about not knowing what n is right now, just keep in mind that n is the number of rectangles you'll make, and it can represent any number you want - for now we'll just keep it as n) Make sure that part makes sense first.

Okay, now the heights can get a bit tricky and it'd be easier if I could draw for you, but I'll try my best to explain.

If you drew the graph of f(x) = x^5 on the interval [0,2], and you decided to cut the graph with 10 rectangles. First your width of each rectangle would be 1/5 (since Δx = (2 - 0)/10 = 2/10 = 1/5) If you drew in your rectangles, can you see that your rectangles height's are dictated by the y - values of the graph? In other words the height of your rectangle will touch the graph of x^5 itself. So, one of the heights would be f(1/5) = (1/5)^5 another would be at f(2/5) = (2/5)^5 another would be at f(3/5) = (3/5)^5, etc until you reach your last rectangle. I recommend drawing this out so you can see this. Label your x - axes with 0, 1/5, 2/5, 3/5, ..., 9/5, 10/5. Draw in 10 rectangles and can you see how the heights of the rectangles are at the y - values of the x - values we just labeled? If you can understand this, then move on...

So, if you cut up n rectangles, you'll have a width of Δx = 2/n (as explained above), and we'll call each x - value: 0, x₁, x₂, x₃, x_{4, ...,} x_n. (Here x_n lands on 2, but I'll just call it x_n just to keep everything consistent).

Now the first rectangle's area will be f(x₁)Δx (where Δx is the width of the first rectangle, and f(x₁) is the height of the same rectangle. As we know area = height * width.)

The second rectangle has an area of f(x₂)Δx (same reason as above)

The third rectangle has an area of f(x₃)Δx

The fourth f(x₄)Δx....get the idea? All the way up to the last rectangle which will have area f(x_n)Δx_n

Thus the total area of all the rectangles added up is:

f(x₁)Δx + f(x₂)Δx + f(x₃)Δx + f(x₄)Δx + ... + f(x_n)Δx

Now, since all the Δx's are the same (i.e. 2/n) I can factor out the Δx and I'd have:

[ f(x₁) + f(x₂) + f(x₃) + f(x₄) + ... + f(x_n) ]Δx

Which using summation notation, I can write as:

n

Σ f(x_i) Δx <-----This is just shorthand for what you see in the previous line above.

i=1

Now, because you want to know the exact area under the curve, you need more rectangles. Thus n→∞.

So, n

Area = lim ∑ f(x_i)Δx <------ Do you see how this just says add up a whole crap ton of rectangles?

n→∞ i = 1

---------

So, for your part (a) remember your x - values are 0, 2/n, 4/n, 6/n, ..., 2n/n. So, x₁ = 2/n; x₂ = 4/n; x₃ = 6/n, etc.

In other words x_i = (2i)/n

f(x_i)=f(2i/n) = (2i/n)⁵ , and as explained above Δx = 2/n (b/c (2 - 0)/n = 2/n)

Putting this into our formula:

n

Area = lim ∑ (2i/n)⁵(2/n)

n→∞ i = 1

Expanding this out, we get:

lim [(2/n)⁵(2/n) + (4/n)⁵(2/n) + (6/n)⁵(2/n) +(8/n)⁵(2/n) + .... + (2n/n)⁵ (2/n)]

n→∞

(b) Now, starting with the line above

lim [(2/n)⁵(2/n) + (4/n)⁵(2/n) + (6/n)⁵(2/n) + (8/n)⁵(2/n) + .... + (2n/n)⁵ (2/n)]

n→∞

As before, notice how you can factor out the 2/n

lim 2/n [(2/n)⁵ + (4/n)⁵ + (6/n)⁵ + (8/n)⁵ + .... + (2n/n)⁵]

n→∞

Do you notice that we can factor out a 2⁵/n⁵ as well, since

(2/n)⁵ + (4/n)⁵ + (6/n)⁵ + (8/n)⁵ + .... + (2n/n)⁵

=(2/n)⁵ + (2(2)/n)⁵ + (2(3)/n)⁵ + (2(4)/n)⁵ + .... + (2(n)/n)⁵

=2⁵/n⁵(1⁵ + 2⁵ + 3⁵ + 4⁵ + ... + n⁵)

Using your formula given in the problem, we can write

lim 2/n {2⁵/n⁵(1⁵ + 2⁵ + 3⁵ + 4⁵ + ... + n⁵)}

n→∞

lim 2⁶/n⁶ [n²(n+1)²(2n² + 2n - 1)/12] <------This replaced 1⁵ + 2⁵ + 3⁵ + 4⁵ + ... + n⁵

n→∞

Finally, notice that since the numerator has a leading coefficient of 2n⁶ (in the brackets - once everything is factored out) and the denominator has a power of 12n⁶, we know that

lim 2⁶(2n⁶ + .... )/12n⁶ = 2⁶(2)/12 (since the highest power on top and bottom are both the same powers)

n→∞

Hence the area = 2⁷ /12= 128/12 = 32/3

Sorry, this was super long, but I felt that if you understood where the formula comes from, it'll help with other problems as well. Hope this helps.

Mr. K