He mentioned that you can do it can calculate the triangle area: What you want is to calculate the entire area under the x² and then you make the difference with the triangle under x². The area under x² is a³/3 The you need to calculate the line that connect the cut with axis x and the point (a, a²). You know that the slope is 2a so you can easily find that the cut with axis x is a/2. Then you use the area of the triangle formula, BxH/2, where B =a-a/2 and H= a², so you get a³/4. Finally you make the difference and you get a³/12
That is such an easy way to do it. As for getting a power of a^3 in the area, when the area is basically a power of 1 on the x-axis and a power of 2 on the y-axis, that is beyond me. Any ideas?
(The case of area under function y=x^n from 0 to a equalling a^(n+1)/(n+1) (remembering that ∫x^ndx = x^(n+1)/(n+1)) is consistent in all functions of the form y=x^n - an example is that the area of a right triangle with equal leg lengths is s²/2, where s is either of the leg lengths: because both b and h are equal, b*h/2 = s*s/2 = s²/2)
I'm not surprised by third power in two-dimensional case if one dimension is square of the other. The bounding rectangle's area is literally a^3. The surprising part is that it's a simple fraction, not "polluted" by lower powers.
Turns out that it's a simple fraction for all functions f(x) = x^n. With the same setup as in the video I got an area of a^(n + 1) times the fraction (n - 1)/(2n(n + 1)). For both n = 2 and n = 3 you happen to get 1/12. For n = 4 and greater its less nice, still a simple fraction.
For n=3, the tangent line at (a, a^3) is y - a^3 = 3a^2(x-a), that gives a root at x = 2a/3. Then, integrating under the curve gives a^4/4. Then we subtract the right triangle formed by the tangent line and, the x axis. So a^4/4 - 1/2 * (a - 2a/3) * a^3. Simplifying gives a^4/4 - a^4/6, then combining fractions gives a^4/12
Yup, it's probably simpler because it's not too bad to solve for the intersection point either, just use slope of tangent line, and then the area under the Parabola is a pretty straight forward calculation
Yeah, in 2 dimensional space we should expect that the area will be in form x*y*something. But since we are dealing with parabola y=x*2, it is not so surprising that the answer is in form a*a^2*something=a^3*something.
I'd just integrate between the limits 0 and a, and then remove the area of the triangle; indefinite integral of x^2 = (x^3)/3 + c apply limits of a and a^2; the integral of x^2 when x = a = (a^3)/3 the integral of x^2 when x = 0 = 0 the integral of x^2 evaluated between a and 0 = (a^3)/3 - 0 Area under parabola between x=a and x=0 = a^3/3 To find the are of the triangle we need to know the gradient; Gradient of parabola = 2x Gradient of parabola at x=a = 2a The height of the triangle = a^2 The base of the triangle = a/2 Using basic formula for area of triangle = 1/2 * a/2 * a^2 Area of triangle = a^3/4 Now we just subtract the area of the triangle from the area under the curve; Area under parabola between x=a and x=0 = a^3/3 Area of triangle = a^3/4 Area of region = a^3/3 - a^3/4 = a^3/12
I would've found the area under x^2, then found where the tangent line intersects the x axis (use tangent line equation), this gives you knowledge on 2 catets of the right triangle, using them calculate the area and subtract it from the area under x^2.
More "illuminating" (in my very modest opinion) is to express it as 1/4 a³/3 - that is: the area between the tangent and the parabola is 1/4 of the area between the parabola and the x axis.
Okay, I really love this kind of video. Cuz obviously everyone else (like me) would go for getting the difference of the integral of parabola with respect to x and the area of a triangle, but this video gives a little new perspective that you can do the same by integrating with respect to y. It's like the calculus version of walking on walls. But instead of "walking" on "walls", it's "integrating" on the "y-axis" :D
It is better to calculate two simple integrals than one cumbersome one.) It is easy to find the equation of the tangent y= 2ax-a^2. It intersects the Ox axis at the point with the abscissa x=a/2. S= ∫(0 to a/2) x^2 dx + ∫(a/2 to a) [x^2 -(2ax-a^2)] dx= =(1/3)*(a/2)^3 +∫(from a/2 to a) (x-a)^2 dx =(1/3)*(a/2)^3 + (1/3)*( a/2)^3 =a^3/12. At the same time, we notice an interesting fact: the areas of the parts are the same.
I thought of that, but I'm comfortable integrating with respect to y and I'd rather deal with 1 integral tbh. One method really isn't that much more difficult than the other, it's just a matter of preference.
I agree, Trying to convert the integrals to reference y at a calc I level is prone to careless mistakes during the conversion process imo. At the same time, doing this conversion successfully does show one's deeper understanding of the subject instead of just trying to always stick to integrals of dx A late Calc I or early Calc II student can easily solve this problem, but the different ways to do it shows their capacity in the subject
I had a similar question to this, but not the same(harder). There was a line (d1) tangent to y=3x^2 at (x,3x^2), a line (d2) passing through (0,0) and (x,3x^2). The area between the line d2 and parabola was defined as A(x), and the area between the x axis, d1 and the parabola was B(x). The question was: Limit as x aproaches 0 of A(x) divided by B(x)
The area was to the third power because one dimension (y) was the other dimension (x) squared. The area is multiplying dimensions, so it’s just adding exponentials
In terms of x: y = x² f'(x) = 2x point-slope tangent line: y - a² = 2a(x - a) → y = - a² + 2ax Finding the x-intercept 0 = - a(a - 2x) → x = a/2 Finally: ∫ x² dx from 0 to a/2 + ∫ (x² - (-a² + 2ax)) from a/2 to a = a³/12
Its from the oxford undergraduate admissions test and its quite time pressured as you have to solve each question in around 3 minutes. Also i think this is one of the first questions in the whole test so they start you off easy
@@mathevengers1131 S=∫(0 to x0) [√(x/4b)-4ax^2] dx, where x0= 1/(4(a^2*b)^(1/3)) is the abscissa of the intersection point of two given parabolas. S= 1/(48ab). In addition, it follows from the equalities y= 4ax^2 and x=4by^2 that a and b have dimension 1/"length", therefore, the answer for the area S should be 1/(ab), not (ab).
so... the area enclosed by the parabola, tangent and x-axis for a point a is a CONSTANT×the integral up until that point? b/c the integral from a to 0 is (1/3)a^3 so the area enclosed is always a quarter of that? this seems very strange, because as a gets very large we would expect the tangent to be almost a vertical line (because the tangent of f(x) is always increasing as x increases) and the area enclosed to then be approximately equal to the integral from a to 0, right??
I was just to check my intuition about this (two parts): a) First he is finding the area from the y-axis to the tangent line crossing a^2 making a “square” (not really a square/rectangle in this video but for a higher value of a it could be thought of as one). Then he is subtracting the area to the parabola in the first quadrant from the y-axis from this square. Correct? b) Would this only give an approximate value? Or are integrals and Riemann sums not the same?
I have a much easier way (mst pen make it seems much harder) the goal erea equal the (integral x^2 bounded from 0 to a) - the erea of triangle with hight equal to a^2 Then the base equal to a/2 (you get it by getting the equation of the tangent line then find the y = 0) so the goal erea = ((a^3/3)-( a^2*a)/(2*2)) = (a^3/3)-(a^3/4) = a^3(1/3-1/4) = a^3/12
I don’t think so. My merch shop is from Teespring and I don’t think I can put the makers there. I do have an Amazon affiliate link to the markers in my descriptions.
@@MeButOnTheInternet I mean we can multiply both side by x to get the equation xe^x = x^3 then we can take the W function of both side to get x = W(x^3) which is even harder to solve
Hum ... Given the stated text of the question, isn't the answer infinite, because it is never stated that x>0, and, because of that, you cannot disregard the area between the parabola and the x-axis on the left side of the graph ? ;)
You're right, but it is not explicitely stated that we don't want in our defined area points that are between the parabola, the x-axis, above the tangent line AND to the left of (0;0). If you want that, you must add some condition to avoid those ... or be clearer about what you mean by "bounded" ;)
When an area is bounded by several lines/curves etc, it means what is on the 'inside' of all of those. Note carefully that the parabola passes through the origin (0, 0), where it meets the x-axis. Here, it effectively 'cuts off' the area on the -ve side of the x-axis.
@@JDC2890 Actually, no. Saying that an area is bounded is saying that the lines in question are the borders of the region you want and, as (0;0) is not an intersection point between borders lines ( tangency points do not work in the same way than intersection points in this regard ), the area between the stated lines includes all the area between the parabola and the x-axis and that is above an tangent line, including any points wiht x=2x-1 ) when the problem is about ( y=0 ^ y >=2x-1 ^ x>=0 ) ...
He mentioned that you can do it can calculate the triangle area:
What you want is to calculate the entire area under the x² and then you make the difference with the triangle under x².
The area under x² is a³/3
The you need to calculate the line that connect the cut with axis x and the point (a, a²). You know that the slope is 2a so you can easily find that the cut with axis x is a/2.
Then you use the area of the triangle formula, BxH/2, where B =a-a/2 and H= a², so you get a³/4.
Finally you make the difference and you get a³/12
👍!!
Brabo
That is such an easy way to do it. As for getting a power of a^3 in the area, when the area is basically a power of 1 on the x-axis and a power of 2 on the y-axis, that is beyond me. Any ideas?
@@stevencarr4002 Integration! The area under y=x² from 0 to a is integral[0,a](x²)dx = (a³/3+c)-(0³/3+c) = a³/3+c-c = a³/3 ✅
(The case of area under function y=x^n from 0 to a equalling a^(n+1)/(n+1) (remembering that ∫x^ndx = x^(n+1)/(n+1)) is consistent in all functions of the form y=x^n - an example is that the area of a right triangle with equal leg lengths is s²/2, where s is either of the leg lengths: because both b and h are equal, b*h/2 = s*s/2 = s²/2)
I'm not surprised by third power in two-dimensional case if one dimension is square of the other. The bounding rectangle's area is literally a^3. The surprising part is that it's a simple fraction, not "polluted" by lower powers.
Isn't it (a^3)/3?
Turns out that it's a simple fraction for all functions f(x) = x^n. With the same setup as in the video I got an area of a^(n + 1) times the fraction (n - 1)/(2n(n + 1)).
For both n = 2 and n = 3 you happen to get 1/12. For n = 4 and greater its less nice, still a simple fraction.
@@Tezhut for n=3 i got (a^4)/4
For n=3, the tangent line at (a, a^3) is y - a^3 = 3a^2(x-a), that gives a root at x = 2a/3.
Then, integrating under the curve gives a^4/4. Then we subtract the right triangle formed by the tangent line and, the x axis. So a^4/4 - 1/2 * (a - 2a/3) * a^3. Simplifying gives a^4/4 - a^4/6, then combining fractions gives a^4/12
cheese
you can just subtract area of triangle from the total area under curve from 0 to a and you get the same answer
1:17
@@timkw I didn't notice that , thx
I did that
Yup, it's probably simpler because it's not too bad to solve for the intersection point either, just use slope of tangent line, and then the area under the Parabola is a pretty straight forward calculation
Love it when people in comments give a much simpler method to solve the problem than the video.
Yeah, in 2 dimensional space we should expect that the area will be in form x*y*something. But since we are dealing with parabola y=x*2, it is not so surprising that the answer is in form a*a^2*something=a^3*something.
I'd just integrate between the limits 0 and a, and then remove the area of the triangle;
indefinite integral of x^2 = (x^3)/3 + c
apply limits of a and a^2;
the integral of x^2 when x = a = (a^3)/3
the integral of x^2 when x = 0 = 0
the integral of x^2 evaluated between a and 0 = (a^3)/3 - 0
Area under parabola between x=a and x=0 = a^3/3
To find the are of the triangle we need to know the gradient;
Gradient of parabola = 2x
Gradient of parabola at x=a = 2a
The height of the triangle = a^2
The base of the triangle = a/2
Using basic formula for area of triangle = 1/2 * a/2 * a^2
Area of triangle = a^3/4
Now we just subtract the area of the triangle from the area under the curve;
Area under parabola between x=a and x=0 = a^3/3
Area of triangle = a^3/4
Area of region = a^3/3 - a^3/4 = a^3/12
I would've found the area under x^2, then found where the tangent line intersects the x axis (use tangent line equation), this gives you knowledge on 2 catets of the right triangle, using them calculate the area and subtract it from the area under x^2.
Facts that what I did
Just solve from 0 to a integral x^2 it is a^3/3 and subtract area of right triangle with height a^2 and base a/2 it is a^3/4 and you get a^3/12😃
YES HE KNOWS 1:17
More "illuminating" (in my very modest opinion) is to express it as 1/4 a³/3 - that is: the area between the tangent and the parabola is 1/4 of the area between the parabola and the x axis.
Okay, I really love this kind of video. Cuz obviously everyone else (like me) would go for getting the difference of the integral of parabola with respect to x and the area of a triangle, but this video gives a little new perspective that you can do the same by integrating with respect to y.
It's like the calculus version of walking on walls. But instead of "walking" on "walls", it's "integrating" on the "y-axis" :D
It is better to calculate two simple integrals than one cumbersome one.)
It is easy to find the equation of the tangent y= 2ax-a^2. It intersects the Ox axis at the point with the abscissa x=a/2.
S= ∫(0 to a/2) x^2 dx + ∫(a/2 to a) [x^2 -(2ax-a^2)] dx=
=(1/3)*(a/2)^3 +∫(from a/2 to a) (x-a)^2 dx =(1/3)*(a/2)^3 + (1/3)*( a/2)^3 =a^3/12.
At the same time, we notice an interesting fact: the areas of the parts are the same.
I thought of that, but I'm comfortable integrating with respect to y and I'd rather deal with 1 integral tbh. One method really isn't that much more difficult than the other, it's just a matter of preference.
I agree,
Trying to convert the integrals to reference y at a calc I level is prone to careless mistakes during the conversion process imo.
At the same time, doing this conversion successfully does show one's deeper understanding of the subject instead of just trying to always stick to integrals of dx
A late Calc I or early Calc II student can easily solve this problem, but the different ways to do it shows their capacity in the subject
I had a similar question to this, but not the same(harder).
There was a line (d1) tangent to y=3x^2 at (x,3x^2), a line (d2) passing through (0,0) and (x,3x^2). The area between the line d2 and parabola was defined as A(x), and the area between the x axis, d1 and the parabola was B(x). The question was:
Limit as x aproaches 0 of A(x) divided by B(x)
The answer is
2. Via (x^3)/2 divided by (x^3)/4
The area was to the third power because one dimension (y) was the other dimension (x) squared. The area is multiplying dimensions, so it’s just adding exponentials
i am so proud of myslf, this is the first time I found the answer by myself by trying it first
The tangent line is (y-a²)=2a(x-a)
Doing y=0 makes x=a/2 hence T=a³/4
Thus A = a³/3 - a³/4 = a³/12
I obtained the area by deducting the area of triangle (0.25a^3)from the area under the curve(a^3/3).
the area is expressed as a^3, because its the product of a tiimes a^2 times some value less than 1
In terms of x:
y = x²
f'(x) = 2x
point-slope tangent line:
y - a² = 2a(x - a)
→ y = - a² + 2ax
Finding the x-intercept
0 = - a(a - 2x) → x = a/2
Finally:
∫ x² dx
from 0 to a/2
+ ∫ (x² - (-a² + 2ax))
from a/2 to a
= a³/12
you could save some integration time by changing the upper limit in the first one to 'a' and dropping the x^2 from the second one
I can’t believe this problem is from Oxford! It’s very easy.
Its from the oxford undergraduate admissions test and its quite time pressured as you have to solve each question in around 3 minutes. Also i think this is one of the first questions in the whole test so they start you off easy
Area between y=4ax^2 and x=4by^2 is 16ab/3
Are you sure?)
It seems to be 1/(48*a*b).
@@Vladimir_Pavlov I am sure
@@mathevengers1131 S=∫(0 to x0) [√(x/4b)-4ax^2] dx, where x0= 1/(4(a^2*b)^(1/3)) is the abscissa of the intersection point of two given parabolas.
S= 1/(48ab).
In addition, it follows from the equalities y= 4ax^2 and x=4by^2 that a and b have dimension
1/"length", therefore, the answer for the area S should be 1/(ab), not (ab).
so... the area enclosed by the parabola, tangent and x-axis for a point a is a CONSTANT×the integral up until that point? b/c the integral from a to 0 is (1/3)a^3 so the area enclosed is always a quarter of that? this seems very strange, because as a gets very large we would expect the tangent to be almost a vertical line (because the tangent of f(x) is always increasing as x increases) and the area enclosed to then be approximately equal to the integral from a to 0, right??
I was just to check my intuition about this (two parts):
a) First he is finding the area from the y-axis to the tangent line crossing a^2 making a “square” (not really a square/rectangle in this video but for a higher value of a it could be thought of as one). Then he is subtracting the area to the parabola in the first quadrant from the y-axis from this square. Correct?
b) Would this only give an approximate value? Or are integrals and Riemann sums not the same?
It gives an exact answer; integrals and Riemann sums are two different things
Can we do double integral?
Yea that would be essentially the same as what I did.
It’s easyer to integrate dx and substract the area of the triangle
I have a much easier way (mst pen make it seems much harder)
the goal erea equal the (integral x^2
bounded from 0 to a) - the erea of triangle with hight equal to a^2
Then the base equal to a/2 (you get it by getting the equation of the tangent line then find the y = 0)
so the goal erea = ((a^3/3)-( a^2*a)/(2*2)) =
(a^3/3)-(a^3/4) = a^3(1/3-1/4) = a^3/12
he knows bro just watch the start of the video
would it work if i just did a double integral to calculate the area?
Why can’t we just integrate x^2 and then subtract the area of the triangle? It’s just x^3 / 3 - a^3 / 2
Hey Steve, just a quick question off topic. Would it be possible for you to sell expo 12 marker packs in your shop?
I don’t think so. My merch shop is from Teespring and I don’t think I can put the makers there. I do have an Amazon affiliate link to the markers in my descriptions.
@@bprpcalculusbasics I will have a look, thank you
very good♥
is it possible algebraicly to solve for x in the question
e^x = x^2
No
No. Numerically, you can get x ≈ - 0.703.
it is using the Lambert W function
@@MeButOnTheInternet
I am not sure
@@MeButOnTheInternet
I mean we can multiply both side by x to get the equation
xe^x = x^3
then we can take the W function of both side to get
x = W(x^3)
which is even harder to solve
The Squeeze (Sandwich) Theorem Math106 Calculus - RUclips
Hum ... Given the stated text of the question, isn't the answer infinite, because it is never stated that x>0, and, because of that, you cannot disregard the area between the parabola and the x-axis on the left side of the graph ? ;)
But the question was finding the area bounded by the parabola, tangent line and the x-axis.Not the area bounded by the parabola and the x-axis.
You're right, but it is not explicitely stated that we don't want in our defined area points that are between the parabola, the x-axis, above the tangent line AND to the left of (0;0). If you want that, you must add some condition to avoid those ... or be clearer about what you mean by "bounded" ;)
When an area is bounded by several lines/curves etc, it means what is on the 'inside' of all of those. Note carefully that the parabola passes through the origin (0, 0), where it meets the x-axis. Here, it effectively 'cuts off' the area on the -ve side of the x-axis.
@@JDC2890 Actually, no. Saying that an area is bounded is saying that the lines in question are the borders of the region you want and, as (0;0) is not an intersection point between borders lines ( tangency points do not work in the same way than intersection points in this regard ), the area between the stated lines includes all the area between the parabola and the x-axis and that is above an tangent line, including any points wiht x=2x-1 ) when the problem is about ( y=0 ^ y >=2x-1 ^ x>=0 ) ...
So triggered you integrated over y instead of x, it seems easier. But the result is the same so it doesn't matter.
Here in sweden the national test in maths had a very similar question (for 11th graders)
excellento
Integrating with respect to y just feels wrong 😂
Easy!😼😼😼
woo, im early!
That Pokeball that's in your hand wants me to grab it and Chuck it into a fire
Hey I might actually watch ya vids then