Edexcel C1 — Question 6 7 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndefinite & Definite Integrals
TypeFind curve from gradient
DifficultyModerate -0.5 This is a straightforward integration question requiring rewriting the square root as a power, integrating term-by-term using standard rules, finding the constant using given conditions, then substituting x=4. It's slightly easier than average as it involves only routine C1 techniques with no conceptual challenges, though the multi-step process and arithmetic with fractions prevents it from being trivial.
Spec1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08d Evaluate definite integrals: between limits
Given that $$\frac{dy}{dx} = 3\sqrt{x} - x^2,$$ and that \(y = \frac{2}{3}\) when \(x = 1\), find the value of \(y\) when \(x = 4\). [7]
AnswerMarks
\(y = \int (3\sqrt{x} - x^2) \, dx\)
\(y = 2x^{\frac{3}{2}} - \frac{1}{3}x^3 + c\)M1 A2
\(x = 1, y = \frac{2}{3} \therefore \frac{2}{3} = 2 - \frac{1}{3} + c\)
\(c = -1\)M1
\(y = 2x^{\frac{3}{2}} - \frac{1}{3}x^3 - 1\)A1
when \(x = 4, y = 2(\sqrt{4})^3 - \frac{1}{4}(4)^3 - 1\)M1
\(y = 16 - 21\frac{1}{3} - 1 = -6\frac{1}{3}\)A1
Total: 7 marks
$y = \int (3\sqrt{x} - x^2) \, dx$ | 

$y = 2x^{\frac{3}{2}} - \frac{1}{3}x^3 + c$ | M1 A2 | 
$x = 1, y = \frac{2}{3} \therefore \frac{2}{3} = 2 - \frac{1}{3} + c$ | 
$c = -1$ | M1 | 
$y = 2x^{\frac{3}{2}} - \frac{1}{3}x^3 - 1$ | A1 | 
when $x = 4, y = 2(\sqrt{4})^3 - \frac{1}{4}(4)^3 - 1$ | M1 | 
$y = 16 - 21\frac{1}{3} - 1 = -6\frac{1}{3}$ | A1 | 
**Total: 7 marks**

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Given that
$$\frac{dy}{dx} = 3\sqrt{x} - x^2,$$
and that $y = \frac{2}{3}$ when $x = 1$, find the value of $y$ when $x = 4$. [7]

\hfill \mbox{\textit{Edexcel C1  Q6 [7]}}
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