Integration (DP IB Applications & Interpretation (AI): SL): Exam Questions

A function $f$ is defined by the equation  $f\left(x\right)=-3x+35$.

Sketch the graph of  $y=f\left(x\right)$ in the interval  $0\le x\le 10$.

Use your sketch from part (a), along with relevant area formulae, to work out the value of the integral

${\int }_1^9(-3x+35)\mathrm{d}x$

You should not use your GDC to find the value of the integral.

The derivative of the function $f$ is given by

$f^{\text{'}}\left(x\right)=\frac{9}{2}{x}^2+7x-2$

and the curve $y=f\left(x\right)$   passes through the point $\left(-3,-\frac{11}{2}\right)$ .

Find an expression for $f$.

A curve $y=f\left(x\right)$  has the gradient function $f^{\text{'}}\left(x\right)=4ax+6$  ,  where $a\in \mathrm{ℝ}$ is a constant.  The diagram below shows part of the curve, with the $x$ and $y$ intercepts labelled and where $\mathrm{V}$ represents the vertex of the curve.

Find

(i) the value of $a$

(ii) the equation of the curve $y=f\left(x\right)$

(iii) the coordinates of $\mathrm{V}$.

Find the area between the curve and the $x$-axis.

A section of the curve with equation   $y=\frac{1}{2}\left(x-1\right)\left(x+5\right)$  is shown below:

The shaded region $S$ in the diagram is bounded by the curve, the $x$-axis and the line  $x=2$.

(i) Write down an integral for the area of the shaded region $\mathrm{S}$.

(ii) Find the area of $\mathrm{S}$.  Give your answer as a fraction.

The shaded region $\mathrm{R}$ in the diagram is bounded on three sides by the curve, the $x$-axis and the $y$-axis.  The boundary on the fourth side is a straight line parallel to the $x$-axis, and that line, the curve and the line  $x=2$  all intersect at a single point.

Find the area of region $\mathrm{R}$. Give your answer as a fraction.

A company is designing a plastic piece for a new game.  The piece is to be in the form of a prism, with a cross-sectional area as indicated by the shaded region $\mathrm{R}$ in the following diagram:

Region $\mathrm{R}$ is bounded, as shown, by the positive $x$- and $y$-axes and the curve with equation $y=\frac{6\left(x-3\right)}{2x-9}$ .  All units are in centimetres.

Using technology, or otherwise, find the coordinates of the points of intersection of the curve with the $x$- and $y$-axes.

The volume of the puzzle piece is to be 30 ${\mathrm{cm}}^3$.

Find the length of the puzzle piece, giving your answer correct to 3 significant figures.

The following diagram shows part of the graph of  $f\left(x\right)=(2x+1)(4x^2-10x+41)$, $x\in \mathrm{ℝ}$ .  The shaded region $R$ is bounded by the $x$-axis, the $y$-axis and the graph of $f$.

(i) Write down an integral for the area of region $\mathrm{R}$.

(ii) Find the area of region $\mathrm{R}$.

ABCD is a parallelogram with vertices $\mathrm{A}(0,0)$, $\mathrm{B}\left(1,\frac{7}{3}\right)$, $\mathrm{C}$ and $\mathrm{D}(a,0)$, as shown in the diagram below.  The area of ABCD is equal to the area of region $\mathrm{R}$ above.

By first finding the value of $a$, the $x$-coordinate of point $\mathrm{D}$, determine the coordinates of point $\mathrm{C}$. The coordinates should be given as exact fractions.

A curve has the equation $y=-x^3+8x^2-13x+6$ .  Consider the area enclosed by the curve and the positive $x$-axis.

Sketch the curve, shading the area indicated above.

Using the trapezoidal rule with 5 strips, determine an approximation for the shaded area.

Explain, using your sketch from part (a), why it is not possible to determine immediately whether your approximation will be an underestimate or an overestimate.

Using integration, determine the exact value of the shaded area.

Find the percentage error of the approximation found in part (b), compared with the exact value.

The shaded region $\mathrm{R}$ in the following diagram is bounded by the $x$-axis, the line  $y=8x-4$  and the curve  $y=-x^3+x^2+10x+8$.

Using technology, or otherwise, find the coordinates of

(i) the point of intersection between the curve and the line

(ii) the point of intersection between the line and the $x$-axis

(iii) the point of intersection between the curve and the $x$-axis that is shown in the diagram.

Show that the area of region $\mathrm{R}$ is equal to exactly $\frac{439}{12}$ units² .  Be sure to show all of your working.

For a particle $\mathrm{P}$ travelling in a straight line, the velocity, $v$ m/s, of the particle at time $t$ seconds is given by the equation

$v\left(t\right)=2t^2-8t+9,t\ge 0$

Sketch the graph of $v\left(t\right)$ in the interval $0\le t\le 5$ .

The distance travelled between times $t_1$ and $t_2$ by a particle moving in a straight line may be found by finding the area beneath the particle’s velocity-time graph between those two times.

Find the distance travelled by the particle $\mathrm{P}$ between the times   $t=1$ and  $t=4.5$.

After analysing several years of company data, a fast food company has determined that the rate of change of its sales figures can be modelled by the equation

$\frac{\mathrm{dM}}{\mathrm{d}x}=-0.068x^3+0.72x^2-0.88x-1.9,0\le x\le 10$

where $\mathrm{M}$ represents the number of meals sold in a week (in thousands of meals sold), and $x$ represents the amount spent on advertising during the preceding week (in thousands of euros).

It is known as well that 5988 meals are sold in a week where 2000 euros had been spent on advertising during the preceding week.

Find an expression for $\mathrm{M}\left(x\right)$.

Find the maximum number of meals that the company can expect to sell in a week, and the amount of money that the company should spend on advertising during the preceding week to bring about that level of sales. Give your answers to the nearest meal sold and the nearest euro, respectively.  Be sure to justify that the value you find is indeed a maximum.

A function $f$ is a piecewise linear function defined by

Sketch the graph of $y=f\left(x\right)$  in the interval $0\le x\le 12$ .

Use your sketch from part (a), along with relevant area formulae, to work out the value of the integral

${\int }_1^{11}f\left(x\right)dx$

The derivative of the function $f$ is given by

$f^{\text{'}}\left(x\right)=-\frac{3}{x^2}+\frac{1}{2}{x}^2-\frac{2}{3}x+2,x>0$

and the curve  $y=f\left(x\right)$  passes through the point $\left(6,\frac{65}{2}\right)$ .

Find an expression for $f$.

A curve  $y=f\left(x\right)$  has the gradient function  $f^{\text{'}}\left(x\right)=ax-1$ .  The diagram below shows part of the curve, with the $x$- and $y$-intercepts labelled.

Find

(i) the value of $a$

(ii) the equation of the curve $y=f\left(x\right)$

(iii) the coordinates of point $\mathrm{M}$.

Find the area of the region enclosed by the curve and the $x$-axis.

Celebrity chef Pepper Bee has opened a new restaurant and is charging diners £630 for a piece of his signature ‘Croesus’ cake.  The chef claims that the price reflects the high cost of the gold foil that is placed on top of each slice of cake, but a suspicious and disgruntled customer has decided to investigate this claim.

 The shaded area in the diagram below shows the shape of the piece of gold foil that is placed on top of each slice of cake:

The shape is that of a rectangle, from which four identical curved sections have been removed.  The rectangle is bounded by the positive $x$- and $y$-axes and the lines  $x=10$  and $y=6$ .  The shape of one of the curved sections in the diagram can be described by the curve with equation

$y=-\frac{1}{4}\left(2x-7\right)\left(2x-13\right)$

All units are given in centimetres.

Given that gold foil costs  $\pounds 0.004788$  per ${\mathrm{mm}}^2$, work out the cost of the gold foil on a piece of Pepper Bee’s Croesus cake.  Give your answer to 2 decimal places.

A company is designing a piece for one of the plastic wargaming models they produce.  The piece is to be in the form of a prism, with a cross-sectional area as indicated by the shaded region $\mathrm{R}$ in the following diagram:

Region $\mathrm{R}$ is bounded, as shown, by the positive $x$-axis and the curve with equation  $y=$. All units are in centimetres.

Given that the model piece will have a volume of 50.3 cm³, find the length of the piece.

The following diagram shows part of the graph of $f\left(x\right)=\frac{1}{2}\left(2x+1\right)\left(x^2-\frac{5}{2}x+8\right),x\in \mathrm{ℝ}$, .  The shaded region $\mathrm{R}$ is bounded by the $x$-axis, the $y$-axis and the graph of $f$.

Find the area of region  $\mathrm{R}$

A trapezoid $\mathrm{ABCD}$ is shown below.

$\left[\mathrm{AB}\right]$ is perpendicular to $\left[\mathrm{AD}\right]$ and parallel to $\left[\mathrm{CD}\right]$. $\mathrm{CD}=\frac{34}{45}$  .  The coordinates of points $\mathrm{A}$, $\mathrm{B}$ and $\mathrm{D}$ are $(0,1)$, $\left(\frac{9}{10},p\right)$ and$\left(\frac{3}{4},\frac{7}{16}\right)$  respectively, where $p>0$ is a constant.

 Given that $\mathrm{ABCD}$ has the same area as the region R above, find the value of $p$, the   $y$-coordinate of point $\mathrm{B}$.

A curve has the equation  $y=x^3-5x^2+2x+8$.

Sketch the curve.

Using the trapezoidal rule with $n=5$, determine an approximation for the integral

${\int }_{-1}^{\frac{3}{2}}(x^3-5x^2+2x+8)dx$

Give your answer as an exact value.

Explain, using your sketch from part (a), why your approximation will be an underestimate.

(i) Find the exact value of the integral from part (b).

(ii) Find the percentage error of the approximation found in part (b), compared with the exact value of the integral.

Explain how you might modify your method in part (b) in order to get a more accurate approximation.

The shaded region  in the following diagram is bounded by the two curves     $y=5x^2-12x+8$ and  $y=-3x^2+10x+29$ .

The two curves intersect at points $\mathrm{A}$ and $\mathrm{B}$ as shown.  $x_A$ and $x_B$ are the $x$-coordinates of points $\mathrm{A}$ and $\mathrm{B}$ respectively.

By setting up and solving an appropriate quadratic equation, find the values of $x_A$ and $x_B$

Find the area of region $\mathrm{R}$ , giving your answer as an exact value.

For a particle $P$ travelling in a straight line, the velocity, $v$ m/s, of the particle at time $t$ seconds is given by the equation

$v\left(t\right)=t^3-15t^2+48t+64,0\le t\le 10$

At time $t_1$ the particle reaches its maximum velocity, while at time $t_2$ the particle comes momentarily to rest.

Find the values of $t_1$and $t_2$, justifying your answers in each case.

The distance travelled between two times by a particle moving in a straight line may be found by finding the area beneath the particle’s velocity-time graph between those two times.

Find

(i) the total distance travelled by the particle $\mathrm{P}$ between times   $t=0$  and  $t=10$ .

(ii) the percentage of that total distance that is covered between times $t_1$ and $t_2$.

Donty is a would-be social media celebrity who is obsessed with the number of ‘likes’ his posts receive.  He hires a statistician to study his social media accounts, and after analysing several years of data she determines that the rate of change of his number of ‘likes’ can be modelled by the equation

$\frac{\mathrm{dL}}{\mathrm{d}x}=-0.164x^3+2.73x^2-12.7x+15.3,0\le x\le 12$

where $\mathrm{L}$ represents the number of likes received on a given day (in thousands of likes), and $x$ represents the amount of new video content Donty uploaded on the preceding day (in hours).  Because of technical limitations, Donty is unable to upload more than 12 hours of new video content on any given day.

It is known as well that 36075 ‘likes’ are received on a day after 5 hours of video content was uploaded the day before

Find the maximum and minimum number of ‘likes’ that Donty can expect to receive in a day, and the corresponding number of hours of new video content that Donty should upload on the preceding day to attain that maximum or minimum. Be sure to justify that the values you find are indeed the maximum and minimum possible.

(i) For the maximum value determined in part (a), calculate the number of likes that are received for each minute of new video content uploaded the preceding day.

(ii) State, with a reason, whether the value calculated in part (b) (i) represents the maximum number of ‘likes per minute of new content’ that Donty is able to achieve.