How many SCSA Mathematics Methods questions cover Integrals?

AusGrader has 114 SCSA Mathematics Methods questions on Integrals, all with instant AI grading and detailed marking feedback.

SCSA (WACE) Mathematics Methods — Integrals Questions & Answers | AusGrader

Q1

2024

QCAA

Paper 1

1 mark

Q1

1 mark

Determine $\int x^4 dx$

A

$4x^3 + c$

B

$5x^5 + c$

C

$\frac{1}{3}x^3 + c$

D

$\frac{1}{5}x^5 + c$

Q3

2021

QCAA

Paper 1

1 mark

Q3

1 mark

Determine $\int 10e^{4x} dx$

A

$\frac{10e^{4x+1}}{4x+1} + c$

B

$40e^{4x} + c$

C

$\frac{5}{2}e^{4x} + c$

D

$2e^{5x} + c$

Q1

2021

SCSA

Paper 1

9 marks

Q1a

3 marks

Differentiate $\frac{3x + 1}{x^3}$ and simplify your answer.

\begin{aligned} \frac{d}{dx}\left(\frac{3x+1}{x^3}\right) &= \frac{x^3(3) - 3x^2(3x+1)}{x^6} \\ &= \frac{3x^3 - 9x^3 - 3x^2}{x^6} \\ &= \frac{-6x - 3}{x^4} \end{aligned}

Marking Criteria

Descriptor	Marks
recognises the need for the quotient rule	1
correctly differentiate the expression	1
simplifies the result	1

Q1b

3 marks

Let $f'(x) = x\ln(2x)$ . Determine a simplified expression for the rate of change of $f'(x)$ .

\begin{aligned} f''(x) &= x \times \frac{2}{2x} + 1 \times \ln(2x) \\ &= 1 + \ln(2x) \end{aligned}

Marking Criteria

Descriptor	Marks
identifies the rate of change as $f''(x)$	1
correctly determines $f''(x)$	1
simplifies the expression for $f''(x)$	1

Q1c

3 marks

Given that $g'(x) = 4e^{2x}$ and $g(1) = 0$ , determine $g(5)$ .

\begin{aligned} g(x) &= 2e^{2x} + c \\ \text{Since } g(1) &= 0, \\ 0 &= 2e^2 + c \\ \therefore c &= -2e^2 \\ \therefore g(x) &= 2e^{2x} - 2e^2 \\ \therefore g(5) &= 2e^{10} - 2e^2 \end{aligned}

Marking Criteria

Descriptor	Marks
states an expression for $g(x)$ , including the constant of integration	1
correctly determines the constant	1
correctly determines $g(5)$ as the final solution	1

Q10

2020

SCSA

Paper 2

7 marks

Q10

Water flows into a bowl at a constant rate. The water level, $h$ , measured in centimetres, increases at a rate given by

$h'(t) = \frac{4t + 1}{2t^2 + t + 1}$

where the time $t$ is measured in seconds.

Q10a

1 mark

Determine the rate that the water level is rising when $t = 2$ seconds.

$h'(2) = \frac{4(2)+1}{2(2)^2+(2)+1} = \frac{9}{11} \text{ cm/s} \quad \{0.818\}$

Marking Criteria

Descriptor	Marks
determines correct rate including units	1

Q10b

2 marks

Explain why $h(t) = \ln(2t^2 + t + 1) + c$ .

$h'(t)$ is of the form $\frac{f'(x)}{f(x)}$ (the numerator is the derivative of the denominator), so the function $h(t)$ is the natural logarithm of the denominator.
Also, $+c$ needs to be included in the function, as any constant could be included here.

Marking Criteria

Descriptor	Marks
states that the numerator is the derivative of the denominator	1
identifies the number $c$ as the constant of integration	1

Q10c

1 mark

Determine the total change in the water level over the first 2 seconds.

$\Delta h = \int_0^2 \frac{4t+1}{2t^2+t+1} dt = \ln(11) \text{ cm} \quad \{2.398\}$

Marking Criteria

Descriptor	Marks
determines total change	1

Q10d

3 marks

The bowl is filled when the water level reaches $\ln(56)$ cm.

If the bowl is initially empty, determine how long it takes for the bowl to be filled.

Let the time taken for the bowl to be filled = $a$ seconds

\begin{align*} \ln(56) &= \int_0^a \frac{4t+1}{2t^2+t+1} dt\\ &= \left[ \ln(2t^2+t+1) \right]_0^a\\ &= \ln(2a^2+a+1)\\ 56 &= 2a^2+a+1\\ a &= 5 \end{align*}

The bowl will take 5 seconds to completely fill.

Marking Criteria

Descriptor	Marks
states a definite integral for depth of water	1
equates definite integral to maximum water level	1
determines time taken	1

Q2

2024

SCSA

Paper 1

10 marks

Q2

An advertising graphic moves across the bottom of a television screen during a sporting game, changing direction to attract viewer attention. The position of the graphic is modelled by

$x(t) = \frac{1}{3}t^3 - 7t^2 + 40t$

where $x$ , in centimetres, is the position of the graphic relative to the left side of the screen, and $t$ , in seconds, is the time from when the graphic first appears on the screen.

The position of the graphic at integer time increments is given in the table below.

$t$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$x(t)$	$0$	$33\frac{1}{3}$	$54\frac{2}{3}$	$66$	$69\frac{1}{3}$	$66\frac{2}{3}$	$60$	$51\frac{1}{3}$

$t$	$8$	$9$	$10$	$11$	$12$	$13$	$14$	$15$
$x(t)$	$42\frac{2}{3}$	$36$	$33\frac{1}{3}$	$36\frac{2}{3}$	$48$	$69\frac{1}{3}$	$102\frac{2}{3}$	$150$

Q2a

2 marks

Determine the velocity of the graphic when it first appears on the screen.

The velocity of the graphic is given by

$\begin{aligned} v(t)&=\dfrac{d}{dt}\left(\dfrac{1}{3}t^3-7t^2+40t\right)\\ &=t^2-14t+40 \end{aligned}$

Hence, when it first appears on the screen

$v(0)=40\text{ cm/s}$

The velocity of the graphic is 40 cm/s to the right of the screen.

Marking Criteria

Descriptor	Marks
determines correct expression for $v(t)$	1
correctly evaluates $v(0)$ to obtain correct answer	1

Q2b

2 marks

Is the graphic initially speeding up or slowing down? Justify your answer.

The acceleration of the graphic is given by

$\begin{aligned} a(t)&=\dfrac{d}{dt}\left(t^2-14t+40\right)\\ &=2t-14 \end{aligned}$

Hence, $a(0)=-14\text{ cm/s}^2.$ The graphic is initially slowing down because the acceleration is in the opposite direction to the velocity.

Marking Criteria

Descriptor	Marks
determines correct expression for $a(t)$	1
concludes that the graphic is slowing down with correct justification (only saying the acceleration is negative is not correct justification)	1

Q2c

3 marks

Evaluate $\int_3^9 v(t)dt$ and explain what this integral represents.

$\begin{aligned} \int_{3}^{9} v(t)\,dt&=x(9)-x(3)\\ &=36-66\\ &=-30\text{ cm} \end{aligned}$

The integral represents the change in displacement/position of the graphic from 3 seconds up to 9 seconds after the graphic appears on screen.

Marking Criteria

Descriptor	Marks
correctly evaluates the integral	1
states that the integral represents a change in displacement/position	1
specifies that the change in displacement is from the 3 second to 9 second marks	1

Q2d

3 marks

Calculate the total distance travelled by the graphic from the time it enters the screen to the time it leaves the screen 15 seconds later.

The graphic is at rest when

$\begin{aligned} v(t)&=0\\ \Rightarrow 0&=t^2-14t+40\\ &=(t-4)(t-10)\\ \Rightarrow t&=4\text{ or }t=10 \end{aligned}$

Hence, the distance $d$ is

$\begin{aligned} d&=|x(4)-x(0)|+|x(10)-x(4)|+|x(15)-x(10)|\\ &=\left|69\dfrac{1}{3}-0\right|+\left|33\dfrac{1}{3}-69\dfrac{1}{3}\right|+\left|150-33\dfrac{1}{3}\right|\\ &=69\dfrac{1}{3}+36+116\dfrac{2}{3}\\ &=222\text{ cm} \end{aligned}$

Marking Criteria

Descriptor	Marks
solves $v(t)=0$ to determine the times at which the graphic is at rest	1
states a correct expression for the distance	1
correctly calculates the total distance travelled	1

SCSA Mathematics Methods Integrals

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