# Thread: [SOLVED] Differentiation &amp; Intergration Question

1. ## [SOLVED] Differentiation &amp; Intergration Question

Question:

The diagrams shows the curve $y = x(x-1)(x-2)$, which crosses the $x-axis$ at the points $O(0,0)$, $A(1,0)$ and $B(2,0)$.

(i) The tangents to the curve at the points A and B meet at the point C. Find the coordinates of C.
(ii) Show by integration that the area of the shaded region $R_1$ is the same as the area of the shaded region $R_2$.

Attempt:

I can't see the point $C$ on the graph!

2. Point C is the intersection between the tangent of the curve at A and the tangent of the curve at B:

Point C is the intersection between the tangent of the curve at A and the tangent of the curve at B:

Can you also explain to me how to get the tangent from the equarion of the curve?

4. Originally Posted by looi76
Question:

The diagrams shows the curve $y = x(x-1)(x-2)$, which crosses the $x-axis$ at the points $O(0,0)$, $A(1,0)$ and $B(2,0)$.

(i) The tangents to the curve at the points A and B meet at the point C. Find the coordinates of C.
First of all get the equations of the tangents.
$y - y_{A}= \left(\frac{dy}{dx}\right)_{A}(x - x_{A})$
$y - y_{B}= \left(\frac{dy}{dx}\right)_{B}(x - x_{B})$
After this solve the above two equations simultaneously to get C. You do this because C lies on both of them.

Originally Posted by looi76
(ii) Show by integration that the area of the shaded region $R_1$ is the same as the area of the shaded region $R_2$.
Where are you stuck here? Try writing the integrals...

Good luck

5. well first you need to find the gradient of the lines AC and BC, remember that differentiating the equation of the curve will give you the gradient at a particular point:

$
\frac{d}{dx}y = x(x-1)(x-2)
$

$
\frac{d}{dx}y = x^3 - 3x^2 +2x
$

$
\frac{dy}{dx} = 3x^2 - 6x +2
$

substituting the coordinates of x given in A(1,0) and B(2,0) you can get the gradient at A and B respectively.

then using the general line equation:

$y=mx+c
$

obtain the line equation for AC and BC

then solving the simultaneous equation(equation of line AC and BC) you can get the coordinates of C

6. Originally Posted by Isomorphism
First of all get the equations of the tangents.
$y - y_{A}= \left(\frac{dy}{dx}\right)_{A}(x - x_{A})$
$y - y_{B}= \left(\frac{dy}{dx}\right)_{B}(x - x_{B})$
After this solve the above two equations simultaneously to get C. You do this because C lies on both of them.

Where are you stuck here? Try writing the integrals...

Good luck
well first you need to find the gradient of the lines AC and BC, remember that differentiating the equation of the curve will give you the gradient at a particular point:

$
\frac{d}{dx}y = x(x-1)(x-2)
$

$
\frac{d}{dx}y = x^3 - 3x^2 +2x
$

$
\frac{dy}{dx} = 3x^2 - 6x +2
$

substituting the coordinates of x given in A(1,0) and B(2,0) you can get the gradient at A and B respectively.

then using the general line equation:

$y=mx+c
$

obtain the line equation for AC and BC

then solving the simultaneous equation(equation of line AC and BC) you can get the coordinates of C

$y = x(x - 1)(x - 2)$
$y = (x^2 - x)(x - 2)$
$y = x^3 - 2x^2 + -x^2 + 2x$
$y = x^3 - 3x^2 + 2x$

$\frac{dy}{dx} = 3x^{3-1} - 3 \times 2x^{2-1} + 2$

$\frac{dy}{dx} = 3x^2 - 6x + 2$

$\frac{dy}{dx}_A | x = 1$███ $\frac{dy}{dx}_A = 3(1)^2 - 6(1) + 2 = -1$

$\frac{dy}{dx}_B | x = 2$███ $\frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = -4$

Equation of tangent to the curve at the point $A(1,0)$
$y - y_1 = m(x - x_1)$
$y - 0 = -1(x - 1)$
$y = -x + 1$

Equation of tangent to the curve at the point $B(2,0)$
$y - y_1 = m(x - x_1)$
$y - 0 = -4(x - 2)$
$y = -4x + 8$

$y + x = 1 \longrightarrow$ equation 1
$y + 4x = 8 \longrightarrow$ equation 2
$-3x = -7$
$x = \frac{7}{3}$

Substitute $x$ in equation $1$
$y + x = 1$
$y + \frac{7}{3} = 1$
$y = 1 - \frac{7}{3}$
$y = -\frac{4}{3}$

$C\left(\frac{7}{3},-\frac{4}{3}\right)$

Is it correct?

7. well you got the idea there but you made a little mistake in calculations

$\frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = 2$

8. Originally Posted by looi76

$y = x(x - 1)(x - 2)$
$y = (x^2 - x)(x - 2)$
$y = x^3 - 2x^2 + -x^2 + 2x$
$y = x^3 - 3x^2 + 2x$

$\frac{dy}{dx} = 3x^{3-1} - 3 \times 2x^{2-1} + 2$

$\frac{dy}{dx} = 3x^2 - 6x + 2$

$\frac{dy}{dx}_A | x = 1$███ $\frac{dy}{dx}_A = 3(1)^2 - 6(1) + 2 = -1$

$\frac{dy}{dx}_B | x = 2$███ $\frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = -4$

Equation of tangent to the curve at the point $A(1,0)$
$y - y_1 = m(x - x_1)$
$y - 0 = -1(x - 1)$
$y = -x + 1$

Equation of tangent to the curve at the point $B(2,0)$
$y - y_1 = m(x - x_1)$
$y - 0 = -4(x - 2)$
$y = -4x + 8$

$y + x = 1 \longrightarrow$ equation 1
$y + 4x = 8 \longrightarrow$ equation 2
$-3x = -7$
$x = \frac{7}{3}$

Substitute $x$ in equation $1$
$y + x = 1$
$y + \frac{7}{3} = 1$
$y = 1 - \frac{7}{3}$
$y = -\frac{4}{3}$

$C\left(\frac{7}{3},-\frac{4}{3}\right)$

Is it correct?
well you got the idea there but you made a little mistake in calculations

$\frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = 2$
$\frac{dy}{dx}_A | x = 1$███ $\frac{dy}{dx}_A = 3(1)^2 - 6(1) + 2 = -1$

$\frac{dy}{dx}_B | x = 2$███ $\frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = 2$

Equation of tangent to the curve at the point $A(1,0)$
$y - y_1 = m(x - x_1)$
$y - 0 = -1(x - 1)$
$y = -x + 1$

Equation of tangent to the curve at the point $B(2,0)$
$y - y_1 = m(x - x_1)$
$y - 0 = 2(x - 2)$
$y = 2x - 4$

$y + x = 1 \longrightarrow$ equation 1
$y + 2x = -4 \longrightarrow$ equation 2
$2x = -4$
$x = \frac{-4}{2}$
$x = -2$

Substitute $x$ in equation $1$
$y + x = 1$
$y + (-2) = 1$
$y = 1 + 2$
$y = 3$

$C(-2,3)$

Is it correct?

9. $y + x = 1 \longrightarrow$ equation 1
$y + 2x = -4 \longrightarrow$ equation 2
$2x = -4$ <======== where did this came from??
$x = \frac{-4}{2}$
$x = -2$

this part is wrong.

$y + x = 1 \longrightarrow$ equation 1
$y + 2x = -4 \longrightarrow$ equation 2
$2x = -4$ <======== where did this came from??
$x = \frac{-4}{2}$
$x = -2$

this part is wrong.
$y + x = 1 \longrightarrow$ equation 1
$y - 2x = -4 \longrightarrow$ equation 2
$-3x = 5$
$x = \frac{5}{-3}$
$x = -\frac{5}{3}$

Substitute $x$ in equation $1$
$y + x = 1$
$y + -\frac{5}{3} = 1$
$y = 1 -\frac{5}{3}$
$y = -\frac{2}{3}$

$C(-\frac{5}{3},-\frac{2}{3})$

Is it correct?

11. lol you missed out on a small negative sign

equation 1
equation 2

you should get 5/3 for x and -2/3 for y

12. Originally Posted by looi76
Question:

The diagrams shows the curve $y = x(x-1)(x-2)$, which crosses the $x-axis$ at the points $O(0,0)$, $A(1,0)$ and $B(2,0)$.

(i) The tangents to the curve at the points A and B meet at the point C. Find the coordinates of C.
(ii) Show by integration that the area of the shaded region $R_1$ is the same as the area of the shaded region $R_2$.

Attempt:

I can't see the point $C$ on the graph!
For part two show that $\int_0^{1}x(x-1)(x-2)dx=\bigg|\int_1^{2}x(x-1)(x-2)dx\bigg|$