A curve has equation \(y = \mathrm{f}(x)\), where \[\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2} - \frac{2}{x^{2}}, \qquad x \gt 0.\] The curve passes through t...

The gradient function is given and the curve itself is wanted, so integration is the required operation: integration undoes differentiation. The point \((1,\, 4)\) is not decoration. Integration produces a whole family of curves differing by a constant, and one point is exactly what is needed to select the single member that passes through it.

(a) Finding \(\mathrm{f}(x)\) [4 marks]

Nothing can be integrated while a variable sits in a denominator, so rewrite the fraction as a negative power first:

\[\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2} - \frac{2}{x^{2}} = 3x^{2} - 2x^{-2} \qquad \textbf{M1}\]

Now integrate term by term, raising each index by one and dividing by the new index:

\[\int 3x^{2}\,\mathrm{d}x = \frac{3x^{3}}{3} = x^{3}, \qquad \int -2x^{-2}\,\mathrm{d}x = \frac{-2x^{-1}}{-1} = +2x^{-1}\] \[y = x^{3} + 2x^{-1} + c \qquad \textbf{M1 A1}\]

The M1 marks are for the method, one for the rewriting and one for raising a power correctly, and they survive a numerical slip; the A1 needs the integrated expression exactly right, including the constant \(c\). The sign of the second term is where this part bites. Dividing \(-2\) by the new index \(-1\) turns the term positive, so \(-\dfrac{2}{x^{2}}\) integrates to \(+\dfrac{2}{x}\). Writing \(-\dfrac{2}{x}\) is the commonest error, and it is fatal to the accuracy marks.

Use the point to find \(c\). Substituting \(x = 1\) and \(y = 4\):

\[1 + 2 + c = 4 \quad \Longrightarrow \quad c = 1, \qquad \text{so } \mathrm{f}(x) = x^{3} + \frac{2}{x} + 1 \qquad \textbf{A1}\]

Omitting \(+c\) altogether is the classic loss: without it there is no equation to solve and the final accuracy mark cannot be earned. Check the answer by differentiating back: \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^{3} + 2x^{-1} + 1\right) = 3x^{2} - 2x^{-2}\), which is the given gradient function, and \(\mathrm{f}(1) = 1 + 2 + 1 = 4\) as required.

(b) The value of \(y\) when \(x = 2\) [1 mark]

Substitute directly into the function just found:

\[\mathrm{f}(2) = 2^{3} + \frac{2}{2} + 1 = 8 + 1 + 1 = 10 \qquad \textbf{B1}\]

This is a B mark, awarded for the correct value alone, though it is allowed as a follow-through from a wrong \(\mathrm{f}(x)\) in part (a). Note that the restriction \(x \gt 0\) in the question matters: the curve has a break at \(x = 0\), and only the branch on the positive side is under consideration.

(c) Evaluating the definite integral and interpreting it [3 marks]

The integrand is the same gradient function, so the antiderivative from part (a) can be reused. In a definite integral the constant cancels between the two limits, so it is conventional to omit it:

\[\int_{1}^{2}\left(3x^{2} - 2x^{-2}\right)\mathrm{d}x = \left[x^{3} + \frac{2}{x}\right]_{1}^{2} \qquad \textbf{M1}\]

The M1 is for substituting the limits into a correct antiderivative, upper value minus lower value:

\[= \left(8 + 1\right) - \left(1 + 2\right) = 9 - 3 = 6 \qquad \textbf{A1}\]

The interpretation carries the final mark, and it must be stated, not implied. Since \(\mathrm{f}\) is an antiderivative of the integrand, the fundamental theorem of calculus makes the definite integral equal to the change in \(\mathrm{f}\):

\[\int_{1}^{2}\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x = \mathrm{f}(2) - \mathrm{f}(1) = 10 - 4 = 6 \qquad \textbf{A1}\]

So the answer is the increase in \(y\) as \(x\) rises from \(1\) to \(2\), which is exactly the value found in part (b) minus the \(y\)-coordinate of the given point. Two remarks are worth making. First, the constant \(c = 1\) genuinely does not affect the result, because it appears in both \(\mathrm{f}(2)\) and \(\mathrm{f}(1)\) and cancels; that is why the integral could not have told us the value of \(c\) in part (a). Second, this integral should not be described as the area under the curve \(y = \mathrm{f}(x)\). It is the area under the gradient curve, and its meaning for the original curve is a change in height.

Examination takeaway. Convert every fractional term to a negative index before integrating, watch the sign produced by dividing by a negative index, and never omit the constant of integration when a boundary condition is supplied. When a question asks how a definite integral of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) relates to earlier answers, the expected reply is the change in \(y\) between the limits.