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A curve, for which \[ \frac{dy}{dx} = 8x^3 + 3, \] passes through the point $(-1, 3)$ - Scottish Highers Maths - Question 12 - 2023

Question 12

$A-curve,-for-which---\[-\frac{dy}{dx}-=-8x^3-+-3,-\]---passes-through-the-point-$(-1,-3)$-Scottish Highers Maths-Question 12-2023.png$

A curve, for which \[ \frac{dy}{dx} = 8x^3 + 3, \] passes through the point $(-1, 3)$. Express $y$ in terms of $x$.

Worked Solution & Example Answer:A curve, for which \[ \frac{dy}{dx} = 8x^3 + 3, \] passes through the point $(-1, 3)$ - Scottish Highers Maths - Question 12 - 2023

Step 1

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Answer

We start by integrating the expression for ( \frac{dy}{dx} ):

[ y = \int (8x^3 + 3) , dx. ]

The integral of (8x^3) is (\frac{8x^4}{4} = 2x^4) and the integral of (3) is (3x), hence:

[ y = 2x^4 + 3x + C, ]

where (C) is the constant of integration.

Step 2

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Answer

Next, we substitute the point ((-1, 3)) into the equation to solve for (C):

[ 3 = 2(-1)^4 + 3(-1) + C. ]

This simplifies to:

[ 3 = 2(1) - 3 + C ] [ 3 = 2 - 3 + C ] [ 3 = -1 + C ] [ C = 4. ]

Step 3

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Answer

Thus, substituting (C = 4) back into the equation for (y), we have:

[ y = 2x^4 + 3x + 4. ]

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