Moderate -0.3 This is a straightforward integration question requiring rewriting the integrand as 2x^(-3/2), integrating to get -4x^(-1/2), applying limits, and solving a simple equation for a. The algebraic manipulation is routine and the question follows a standard template for AS-level area problems.
6 A curve has equation \(y = \frac { 2 } { x \sqrt { x } }\)
\includegraphics[max width=\textwidth, alt={}, center]{b45dc98e-1699-47c9-9228-5abe0e5c9195-05_508_549_420_744}
The region enclosed between the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = a\) has area 3 units.
Given that \(a > 1\), find the value of \(a\).
Fully justify your answer.
\(\frac{2}{x\sqrt{x}} = 2x^{-\frac{3}{2}}\); expresses as \(x^{-\frac{3}{2}}\) or \(x^{-1.5}\) or \(x^{-1\frac{1}{2}}\)
M1 (AO 1.1a)
PI by correct integration; condone incorrect use of '2'
\(3 = \int_1^a 2x^{-\frac{3}{2}}\,dx = \left[-4x^{-\frac{1}{2}}\right]_1^a\); carries out correct integration to obtain \(-4x^{-\frac{1}{2}}\) OE
A1 (AO 1.1b)
Forms equation by equating 3; PI by correct integral \(\left[-4x^{-\frac{1}{2}}\right]_1^a\), original expression as integral with powers \(\int 2x^{-\frac{3}{2}}\,dx\), or original expression as integral \(\int \frac{2}{x\sqrt{x}}\,dx\), with limits 1 and \(a\), and expression after integration and using limits
M1 (AO 3.1a)
If assuming \(a=16\) and trying to verify scores M1A1M1 max
Substitutes \(x=1\) as lower limit and \(x=a\) as upper limit; \(3 = -4a^{-\frac{1}{2}} + 4\), so \(a^{-\frac{1}{2}} = \frac{1}{4}\)
M1 (AO 1.1a)
\(a = 16\) CAO
A1 (AO 1.1b)
NB \(a=16\) with no justification scores 0/5
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2}{x\sqrt{x}} = 2x^{-\frac{3}{2}}$; expresses as $x^{-\frac{3}{2}}$ or $x^{-1.5}$ or $x^{-1\frac{1}{2}}$ | M1 (AO 1.1a) | PI by correct integration; condone incorrect use of '2' |
| $3 = \int_1^a 2x^{-\frac{3}{2}}\,dx = \left[-4x^{-\frac{1}{2}}\right]_1^a$; carries out correct integration to obtain $-4x^{-\frac{1}{2}}$ OE | A1 (AO 1.1b) | |
| Forms equation by equating 3; PI by correct integral $\left[-4x^{-\frac{1}{2}}\right]_1^a$, original expression as integral with powers $\int 2x^{-\frac{3}{2}}\,dx$, or original expression as integral $\int \frac{2}{x\sqrt{x}}\,dx$, with limits 1 and $a$, and expression after integration and using limits | M1 (AO 3.1a) | If assuming $a=16$ and trying to verify scores M1A1M1 max |
| Substitutes $x=1$ as lower limit and $x=a$ as upper limit; $3 = -4a^{-\frac{1}{2}} + 4$, so $a^{-\frac{1}{2}} = \frac{1}{4}$ | M1 (AO 1.1a) | |
| $a = 16$ CAO | A1 (AO 1.1b) | NB $a=16$ with no justification scores 0/5 |
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6 A curve has equation $y = \frac { 2 } { x \sqrt { x } }$\\
\includegraphics[max width=\textwidth, alt={}, center]{b45dc98e-1699-47c9-9228-5abe0e5c9195-05_508_549_420_744}
The region enclosed between the curve, the $x$-axis and the lines $x = 1$ and $x = a$ has area 3 units.
Given that $a > 1$, find the value of $a$.\\
Fully justify your answer.\\
\hfill \mbox{\textit{AQA AS Paper 2 2019 Q6 [5]}}