1. ## Define Integral Problem

Diagram shows the straight line $y = -x + k$ touching the curve $y = 3x - x^2$ at point $A$ .

a)Find the value of $k$

b)the coordinates of point $A$ .

Diagram shows the straight line $y = -x + k$ touching the curve $y = 3x - x^2$ at point $A$ .

a)Find the value of $k$

b)the coordinates of point $A$ .
Let the x-coordinate of A be x = a. Now use the fact that 3 - 2a = -1 (why?). Therefore y = -a + k (why?). Therefore -a + k = 3a - a^2 (why?). Hence solve for k.

3. i dont understand..

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

how can u get $\frac{dy}{dx} = 3 - 2x = -1?$

4. Hi!

i dont understand..

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

how can u get $\frac{dy}{dx} = 3 - 2x = -1?$
The point A has the x-coordinate a.

You do have the straight line $y_1 = -x+k$and the curve $3x-x^2$.

you know the straight line touches the curve at point A with the x-coordinate a

So that means at x = a (point A) the straight line has the same increase at point A as the curve.

That means

$y_1 ' = y_2 '$

$\frac{dy_1}{dx}= \frac{dy_2}{dx}$

more precisely:

$\frac{d}{dx}(-x+k) = \frac{d}{dx}(3x-x^2)$

$-1 = \frac{d}{dx}(3x-x^2)$

$-1 = 3-2x$

5. thus, the value of $k = 2$ and the coordinates of point $A = (2,0)$ right ?

6. if the question want the area of shaded region

at the curve $y = 3x-x^2$ the $x-intercept$ is

$0=3x - x^2$

$x(3 - x)=0$

$x = 3$

hence, $y=-x^2 + 3x$

$\int_{2}^{3} {y} \ dx$

$\int_{2}^{3} (-x^2 + 3x) dx$

$\left[ \frac{-x^3}{3} + \frac {3x^2}{2} \right]_{2}^{3}$

$\left[\frac{-(3)^3}{3} + \frac{3(3)^2}{2} \right] - \left[\frac{-(2)^3}{3} + \frac{3(2)^2}{2} \right]$

$\left[-9 + \frac{27}{2} \right] - \left[\frac{-8}{3} + 6 \right]$

$=\frac{7}{6} unit^2$

am i right?

thus, the value of $k = 2$ and the coordinates of point $A = (2,0)$ right ?
No! It is A=(2,2) and k =4.

As you know we shall solve

3 - 2a = -1

solving for a :

-2a = -1 -3

-2a = -4

a = 2

now substitute a = 2 in -a + k = 3a - a^2 (as mr fantastic was saying)

-2+k = 3*2 - 2^2

-2+k = 6-4

-2+k = 2

k = 2+2 = 4

So the straight line is y = -x+4

now substitute x = 2 and you have

y = -2+4 =2

So A = (2,2)

if the question want the area of shaded region

at the curve $y = 3x-x^2$ the $x-intercept$ is

$0=3x - x^2$

$x(3 - x)=0$

$x = 3$

hence, $y=-x^2 + 3x$

$\int_{2}^{3} {y} \ dx$

$\int_{2}^{3} (-x^2 + 3x) dx$

$\left[ \frac{-x^3}{3} + \frac {3x^2}{2} \right]_{2}^{3}$

$\left[\frac{-(3)^3}{3} + \frac{3(3)^2}{2} \right] - \left[\frac{-(2)^3}{3} + \frac{3(2)^2}{2} \right]$

$\left[-9 + \frac{27}{2} \right] - \left[\frac{-8}{3} + 6 \right]$

$=\frac{7}{6} unit^2$

am i right?
Yes, you are! GJ

8. Ok..I very understand sir..Thanks!

lastly,if the question want the volume generated,in term of $\pi$,
when the region bounded by the curve $y=3x - x^2$ and the $x-axis$ is revolved through $360^0$ about the $x - axis$

i get,

$=\int_{2}^{3} \pi{y^2} dx$

$=\int_{2}^{3} \pi(3x-x^2)^2 dx$

$=\pi\int_{2}^{3} (9x^2-6x^3+x^4) dx$

$=\pi \left[\frac{9x^3}{3}-\frac{6x^4}{4}+\frac{x^5}{5} \right]_{2}^{3}$

$=\pi\left[\frac{9(3)^3}{3}-\frac{6(3)^4}{4}+\frac{(3)^5}{5} \right] - \left[\frac{9(2)^3}{3}-\frac{6(2)^4}{4}+\frac{(2)^5}{5} \right]$

$=\pi\left(81-\frac{243}{2} + \frac{243}{5} \right) - \left(24-24 + \frac{32}{5}\right)$

$=\frac{17}{10} \pi unit^3$

right?

lastly,if the question want the volume generated,in term of $\pi$,
when the region bounded by the curve $y=3x - x^2$ and the $x-axis$ is revolved through $360^0$ about the $x - axis$

i get,

$=\int_{2}^{3} \pi{y^2} dx$
okay, that's the formula for the volume generated between x=2 and x=3

$=\frac{17}{10} \pi unit^3$

right?
It sure is.

10. thanks very much sir..sorry for wasting your time..but i really appreciate with you..thanks!

11. You're welcome!