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 \((-1, 3)\).
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
Step 1: Integrate the given expression
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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
Step 2: Substitute the point into the equation
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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
Step 3: State the final expression for y
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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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