Відео

Feliz año nuevo desde Nápoles, Italia.

Quick way: Height of shaded area s is given by subtraction s=6x-xx-(xx-2x)=8x-2xx Set 8x-2xx=0 to find limits 2x(4-x)=0 giving limits at x=0, x=4 Integrate s.dx to give 4xx-2xxx/3+c Apply limits 4*16-2*64/3= 64(1-2/3)=64/3=21.33333

5/6✖️64が正解じゃないですかね？

3x/4+2x+4x/3=10 9x+24x+16x=120 49x=120 x=120/49 x²=14400/2401 Area=28800/2401=11.995

Feliz Año!!!🎉🎉🎉

Genial, gracias

El triángulo que los contiene es tipo 3/4/5→ Si "a" es el lado de los cuadrados: 3a/4 +2a +4a/3 =10→ a=120/49→Área de cada cuadrado =a² =14400/2401 ≈ 6 u². Gracias y un saludo cordial.

👍💯🎄Happy New year!

This problem can be solved by integrating from 0 to 1 then from 1 to 2^1/2 in the y direction (dy). The first integral is from 0 to 1 of y^1/2 dy which yields 4/3 the second integral is from 1 to 2^1/2 of (2-Y^2)^1/2 dy, this integral is [arcsin(1/2^1/2 y) + 1/2sin(2arcsin(1/2^1/2 y) this yields pi/2-1. Adding 4/3 from the first integral gives the final answer of pi/2+1/3. Remember when integrating a square root, only one side of the square root (positive side) is considered so you need to multiply by 2 to include both sides.

Acho que você se precipitou no momento de achar os pontos de Corte. Acertou por pura coincidência.

Hola Creo que hay un Error en la solucion, la integral de 6x-x^2 es negativa entre 0 y 2. Por lo tanto As = integral(x^2-2x;0;4) + integral(6x-x^2 entre 0;2) - integral(6x-x^2 entre 2;4)

Me encanto ver tu metodo, yo me enrolle un poco mas al desconocer conceptos de geometria analitica. Utilice puras integrales para resolverlo, lo que hice fue en la ecuacion de la elipse despejar y de modo que me quedara una funcion a integrar (colocandole el - ya que solo me interesa de la parte de abajo), utilice los mismos limites de integracion con los puntos de corte que tenian en y=0. Despues de integrar solo saque la diferencia entre ambas areas y quedo el mismo resultado. Es lo que me encanta de las matematicas, de como poder llegar a un mismo resultado con distintos metodos. cx

Hola, ¿no seria mas directo, integrar la curva del círculo X^2 + Y^2 = 2, entre los puntos -1, +1 y restar el area bajo la curva Y=X^2, que calculaste como A2=2/3?

Wrong. The func y=x^2-2x has Negative square between points x=0 & x=2 (area with the negative sign). You have to transform the form of functions by downing the x-axis to the tip of y=x^2-2x, which is below the x-axis (just to add the delta-y to both of them). And only after that you can calculate the area by the integration of func subtractions, because now they both will have positive areas above the x axis. Actually, it works for this example, but the common solution for the common case is to transform functions by downing x-axis. Especially, if you have a system of complex functions, not just two quadratic functions, for example, trigonometric functions...

This one made my brain cells feeling alive after 4 days of zero sugar. Thx

Ohhh my god ya made this prob wayyy more complicated than it needs to be, including in the integration. Easy way: 2 * integral of [sqrt( 2-x^2) - x^2] from 0 to 1 Why: if integrating can do so from top to bottom…. Literally take the top function minus the bottom function. Can integrate from 0 to 1 because the integration is symmetric. Just multiply by 2 to double the area. Why? B/C ….

Analicemos el gráfico: Vemos la simetría respecto a Y, el círculo es de radio √2, el punto de corte en Y+ es (1,1), si unimos el origen con este punto, notaremos que se forma un ángulo de π/4 respecto a Y. El área a calcular es: A=2*[sector circular (áng π/4)+área de [0,1] de (y=x)-(y=x^2), que es una integral directa] A=2*[π/4+(1/2-1/3)] A=π/2+1/3 Saludos.

Muito boa a explicação!

x^2+x^2× x^2 × x^2= 2 X^2+x^6=2 ❤❤

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Mas mejor

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,Very curiuos task and its solvation as well! Possibilities of integrals are indeed unlimited.💯👍All the best to you, Mr Cristian.:)) Sorry, I don't get Spanish ..

Good. Now find the radius of the largest circle that lies between y=x^2 and y=4

4,2

30 :)

Nice task. I admire integrals. Thank you, Mr Christian.🎄👍💯Merry Christmas!

A=integral de( 2_x^2)^1/2 meno integral de x^2 tra -1 e +1

There's a easier solution: the quarter of circle area is 2*π/4 and +2*integral(x-x^2)|[0,1] = π/2+1/3 Unnecessary dividing area to too many parts.

Explicación muy confusa!

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Se enreda tanto en tratar de explicar lo que se supone ya se sabe que sus videos son aburridos y obsoletos. Vaya al grano directamente y menos bla bla bla que nada más se enreda y nos enreda

Is it x^2 - √2-x^2

The first part where both equations are equated to zero is not really necessary since we are not interested on where the lines intersect the x-axis. We are interested only on which points do the two lines intersect each other (to get the limits). So the first step could be just directly equating the two equations together as equal.

Maravillosamente explicado. Gracias

Why do I feel nostalgic seeing this? In a world of double and triple integrals, volumes under surfaces, inertia, centers or mass, radius of gyration and total charge, this truly was one hell of a time when I first learned this. Crazy how far we can grow in so little time.

Muy bueno, interesante hacerlo solamente con integrales. Saludos.

sustituyendo x2 por y nos queda y+y2=2 es lo mismo? (nos ahorramos la u)

Por que restan las funciones

Circle and parobol intersect at the abscissa 1 (and -1. Be I the integral from 0 to 1 of sqrt(2 - x^2). dx, We note x = sqrt(2).sin(t) and dx = sqrt(2).cos(t).dt. Then I = the integral from 0 to Pi/4 of 2.(cos(t))^2.dt, or the integral from 0 to Pi/4 of (1 + cos(2.t)).dt, so it is [t +(sin(2.t)/2] between 0 and Pi/4, so I = Pi/4 + 1/2 Be J the integral from 0 to 1 of x^2.dx, J = 1/3 (evident). Finally the area we are surching is 2.(I - J) = 2.(Pi/4 + 1/2 - 1/3) = Pi/2 + 1/3.

Excelente explicación, gracias.

I ussually use for similar chase with autocad program 😂

👏🏼👏🏼👏🏼😉 Adorei! Amo Cálculo ❤

A EQUAÇÃO DO CÍRCULO NÃO É UMA FUNÇÃO. POR ISSO A A INTEGRAL NÃO SE APLICA. MAS É UM ÓTIMO EXERCÍCIO.

Brutal explicación

No need to calculate! C and D are unreasonable. A is half of the square, so A is also unreasonable! The answer is B only!

Too complex The length of square is 2r+4, so your x-r=2, your x-3=r-1 Then r^2=4+r^2-2r+1, so r=2.5, then you can know the area. Is it easier than yours?

Interesante, magnífico!!!

Muy buena explicación mas claro que el agua no puede haber. Me recuerda mis clases de ingeniería de geometría analítica👍👍

Your method is too complex. From circle side, the area is (1/4)*pi*(2^0.5)+2*1*1*0.5=(1/2)*pi+1 From y=x^2 side, the area is integral x from -1 to 1, so area=2/3 So, the answer is (1/2)*pi +1 - 2/3=(1/2)*pi +1/3