Prime Newtons has done it again! Bravo! 🎉😊

Prevideo Prediction: If I am remembering my calculus correctly, you would take the integral from 0 to 1 of x^2-x^3 then add the integral from 1 to 2 of x^3-x^2.

Well I noticed the intersection point and did two integrals the same way (x^3-x^2). When the first one turned out to be negative area, I just took the absolute value and added it to the second result. The key of course is as you said, recognizing there is such a point within the range.

Excelent explain!!!! How always......

Right off the bat I can say that this one can trick many people.

Solution: Given, that x³ = x² in x = 1, the two curves intersect there. As such, you need to calculate the area between 0 and 1, as well as the area between 1 and 2. After that, you need to add the absolute values of both to get the desired area. Alternatively, you can analyze, which curve is the bigger in which section and multiply the integral of the flipped section by -1. The latter is relatively simple, as (1/a)³ = 1/a³ < 1/a² = (1/a)². Therefore x³ < x² for 0 < x < 1. It is flipped, so x³ > x², for 1 < x < 2. So the desired area is -1 times the integral of x³ - x² between 0 and 1, plus the integral of x³ - x² between 1 and 2. The base function of x³ - x² is x⁴/4 - x³/3 + c Therefore the area is: A = -1 * ((1⁴/4 - 1³/3) - (0⁴/4 - 0³/3)) + (2⁴/4 - 2³/3) - (1⁴/4 - 1³/3) A = -1 * (1/4 - 1/3) + (16/4 - 8/3) - (1/4 - 1/3) A = -1 * (3/12 - 4/12) + (12/3 - 8/3) - (3/12 - 4/12) A = -1 * (-1/12) + 4/3 - (-1/12) A = 1/12 + 16/12 + 1/12 A = 18/12 = 3/2

Nice explaination! The tempting incorrect answer of integral(from x=0 to x=2 of x^3-x^2) gives 4-8/3=4/3.

I wonder if there's any way to set this up so the right answer comes out automatically, like maybe if you square and then square root the integrand. Of course, it would be helpful only if you're trying to get the area, which isn't normally the case when you do a definite integral. It's probably simpler just to use your brain and intelligently assess the integral, most of the time.

take the absolute value of each integal keeping the limits in order

We had a similar question. It was to find the equation of the curve that bisects the area between y = 9x² and y = 16x². I'd say that was merely easy was us. But I want you to propose a unique solution to the question.

This reminds me of the ones where you have regions under the x axis, and the question asks for the area. alas the integral per se will not give you the right answer, you have to be careful with those :3

For a jee aspirant its easy 😄

Those who stop learning, stop living...🥰🥰... hits anyone hardly.

If x³ was set on top you would get 4÷3(close but no there) and if x² was set on top then it would be -4÷3(very far from the answer)

6:37 o.O :)

Near the end you had 2 + 4 - 1/2, then you had 2 - 1/2. What happened to the 4??

I refuse to let my answer be wrong! First find the intersection points to be x=0 and x=1. Now evaluate the definite integrals for both functions from 0 to 1. You get 1/4 and 1/3. The area between them is 4/12-3/12=1/12. Now evaluate the definite integrals from 1 to 2. You get 15/4 and 7/3. The difference is 45/12-28/12=17/12. The final answer is 1/12+17/12=18/12=1.5.

Too much work. I take the easy way out by just finding when the two functions are equal, which is when x=0 and x=1. Then I compute |int (x^3-x^2)dx from 0 to 1| + |int (x^3-x^2)dx from 1 to 2|, bypassing all the work to see which function is on top and where it is on top.