# Question 6:

## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int_{-1}^{3}(a+bx)\,dx [=1]$ or trapezium drawn | M1 | |
| $\left[ax + \frac{bx^2}{2}\right]_{-1}^{3} [=1]$ oe or $\frac{3-(-1)}{2}((a-b)+(a+3b))[=1]$ | A1 | |
| $\left[3a + \frac{9b}{2}\right] - \left[-a + \frac{b}{2}\right] = 1$ oe or $\frac{4}{2}(2a+2b)=1 \Rightarrow 4a+4b=1$ | A1* | cso |

## Part (b)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int_{-1}^{3} ax^2 + bx^3\,dx = \left[\frac{ax^3}{3} + \frac{bx^4}{4}\right]_{-1}^{3}$ | M1A1 | |
| $\left[\frac{27a}{3} + \frac{81b}{4}\right] - \left[-\frac{a}{3} + \frac{b}{4}\right] = \frac{17}{5}$ | dM1 | |
| $\frac{28}{3}a + 20\left(\frac{1-4a}{4}\right) = \frac{17}{5}$ | M1 A1 | |

## Part (b)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $-\frac{32}{3}a = -\frac{8}{5}$ oe or $\frac{28}{3}\left(\frac{1-4b}{4}\right) + 20b = \frac{17}{5}$ | M1 | |
| $b = \frac{1-4\times0.15}{4} \Rightarrow b = 0.1$ | A1* | cso |

## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct trapezium graph from $x=-1$ to $x=3$, with $f(-1)=0.05$ and $f(3)=0.45$ | M1 A1 | |

## Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\left[\text{"0.15"}k + \frac{0.1k^2}{2}\right] - \left[-\text{"0.15"} + \frac{0.1}{2}\right] = 0.2$ or $\left[0.45 + \frac{0.9}{2}\right] - \left[\text{"0.15"}k + \frac{0.1k^2}{2}\right] = 0.8$ | M1 | |
| or $\frac{1}{2}(k+1)(0.05+0.1k+0.15)=0.2$ or $\frac{1}{2}(3-k)(0.15+0.1k+0.45)=0.8$ | | |
| $0.05k^2 + 0.15k - 0.1 = 0$ | A1 | |
| $k = \frac{-0.15 \pm \sqrt{0.15^2 - 4\times0.05\times(-0.1)}}{2\times0.05}$ | M1 | |
| $= 0.56155...$ awrt **0.562** | A1 | |

# Mark Scheme Extraction

## Question (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempt to integrate $a + bx$ with either $a \rightarrow ax$ or $x \rightarrow x^2$ | **M1** | Ignore limits; or trapezium drawn with parallel sides correct in terms of $a$ and $b$ (may be implied by correct area of trapezium) |
| Correct integration or correct area of trapezium | **A1** | |
| Correct limits substituted or correct area equated to 1, leading to final given answer | **A1\*** | cso |

---

## Question (b)(i):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempt to integrate $ax^2 + bx^3$ with either $x^2 \rightarrow x^3$ or $x^3 \rightarrow x^4$ | **M1** | Ignore limits |
| Correct integration | **A1** | |
| Substituting correct limits and equating to $\frac{17}{5}$ | **dM1** | Dependent on previous M1 |
| Substituting $4b = 1 - 4a$ | **M1** | oe |
| A correct equation | **A1** | |

---

## Question (b)(ii):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Solving equation in $a$ of the form $na = c$ where $n \neq 1$, or a correct equation in terms of $b$ | **M1** | M0 for $a = 0.15$ without working, or for using $b = 0.1$ in $4a + 4b = 1$ to find $a$ |
| Correct unsimplified expression for $b$ leading to given answer $b = 0.1$ | **A1\*** | Must come from correct working |

---

## Question (c):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct shape: straight line with positive gradient, above $x$-axis, between $x = -1$ and $x = 3$ | **M1** | Ignore graph before $-1$ and after $3$ |
| Both correct $x$-axis labels $-1$ and $3$, and at least 1 correct $y$-axis label from $0.05$, $0.15$ or $0.45$ | **A1** | Ignore graph before $-1$ and after $3$ |

---

## Question (d):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct equation using integration or area; need not be simplified | **M1** | Use of limit $k+1$ instead of $k$ in integration is M0 |
| Correct 3-term quadratic | **A1** | oe |
| Correct method to solve their 3-term quadratic, or awrt $0.562$ or awrt $-3.56$ | **M1** | |
| awrt $0.562$, i.e. $\dfrac{\sqrt{17}-3}{2}$, with other solutions eliminated if given | **A1** | |