# Integration help (1 Viewer)

#### ALOZZZ

##### New Member
Hey guys i am having trouble with the following questions:
1) given d2y/dx2=8 and dy/dx=0 and y=3 when x=1, find the equation of y in terms of x.
2) The tangent to a curve with d2y/dx2= 2x-4 makes an angle of inclination of 135 with the x-axis at the point (2,-4). Find the equation
3) A function has a tangent parallel to the line 4x-y-2=0 at the point (0,-2), and f''(x)=12x^2-6x+4. Find the equation of the function
4) A curve has d2y/dx2=6 and a tangent at (-1,3) is perpendicular to the line 2x+4y-3=0. Find the equation of the curve.

#### A1La5

##### New Member
1.

$\bg_white \frac{\mathrm{d^2y} }{\mathrm{d} x^2} = 8$

Integrating, we get:

$\bg_white \frac{\mathrm{dy} }{\mathrm{d} x} = 8x + C_1$

Using the information from the question, $\bg_white \frac{\mathrm{dy} }{\mathrm{dx}$ $\bg_white = 0$ when $\bg_white x = 1$
$\bg_white 8(1) + C_1 = 0$
$\bg_white C_1 = -8$
$\bg_white \frac{\mathrm{dy} }{\mathrm{dx}$ $\bg_white = 8x - 8$

Further integrating the function:
$\bg_white y = 4x^2 -8x + C_2$
Using the information from the question, $\bg_white y = 3$ when $\bg_white x = 1$.
$\bg_white 3 = 4(1)^2 -8(1) + C_2$
$\bg_white C_2 = 7$

Hence, the equation of the curve is: $\bg_white 4x^2 -8x + 7$.

Edit: Renamed constants.

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#### A1La5

##### New Member
4.
$\bg_white \frac{\mathrm{d^2y} }{\mathrm{d} x^2} = 6$
Integrating, we get:

$\bg_white \frac{\mathrm{dy} }{\mathrm{d} x} = 6x + C_1$
In order to find the value of this constant , we need to understand that the gradient of the tangent at (-1,3) is perpendicular to the gradient of the line $\bg_white 2x + 4y - 3 = 0$.
Rearranging: $\bg_white y = \frac{-1}{2}x + \frac{3}{4}$

So, the gradient of the tangent is $\bg_white 2$, as the product of perpendicular gradients is $\bg_white -1$.

Substituting this into $\bg_white \frac{\mathrm{dy} }{\mathrm{d} x} = 6x + C_1$:
$\bg_white 2 = 6(-1) + C_1$
$\bg_white C_1 = 8$
$\bg_white \frac{\mathrm{dy} }{\mathrm{d} x} = 6x + 8$

Integrating again, we get:
$\bg_white y = 3x^2 +8x + C_2$
The point (-1,3) is on this curve. Using this information in the equation:

$\bg_white 3 = 3(-1)^2 + 8(-1) + C_2$
$\bg_white C_2 = 8$

Hence, the equation of the curve is: $\bg_white 3x^2 +8x + 8$

Edit: Renamed constants.

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#### A1La5

##### New Member
Considering no one's tried to help you since I last posted I'll attempt 2 as it helps both of us.

$\bg_white \frac{\mathrm{d^2y} }{\mathrm{dx^2} } = 2x-4$

Integrating:

$\bg_white \frac{\mathrm{dy} }{\mathrm{dx} } = x^2-4x+C_1$

We can source the value of the gradient function at $\bg_white x = 2$ through knowing the angle of inclination.

$\bg_white \theta$ in this case is 135 degrees. Using the relationship above, the gradient function = -1.

Plugging that into the above equation and solving for C:
$\bg_white -1 = (2)^2-4(2)+C_1$
$\bg_white C_1 = 3$
$\bg_white \frac{\mathrm{dy} }{\mathrm{dx} } = x^2-4x+3$

Integrating again:
$\bg_white y = \frac{x^3}{3}-2x^2 +3x + C_2$
Knowing that the point $\bg_white (2,-4)$ is on the curve, it just now boils down to simple substitution.

$\bg_white -4 = \frac{(2)^3}{3}-2(2)^2 +3(2) + C_2$
$\bg_white C_2 = \frac{-14}{3}$

The equation of the curve is: $\bg_white y = \frac{x^3}{3}-2x^2 +3x - 14/3$

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#### Qeru

##### Well-Known Member
Just a pedantic correction, it may be better to name your constants $\bg_white C_1$ and $\bg_white C_2$ rather than both $\bg_white C$ as this implies they are both the same.

#### A1La5

##### New Member
Just a pedantic correction, it may be better to name your constants $\bg_white C_1$ and $\bg_white C_2$ rather than both $\bg_white C$ as this implies they are both the same.
Trust me, I would've if I knew how to. I'm still pretty new to this whole LaTeX business. Thanks though.

#### Qeru

##### Well-Known Member
Trust me, I would've if I knew how to. I'm still pretty new to this whole LaTeX business. Thanks though.
Its quite easy C_n outputs $\bg_white C_n$