IB DP Maths Topic 6.5 Definite integrals HL Paper 2

Question

A function is defined by \(f(x) = {x^2} + 2,{\text{ }}x \ge 0\). A region \(R\) is enclosed by \(y = f(x)\),the \(y\)-axis and the line \(y = 4\).

a.(i)     Express the area of the region \(R\) as an integral with respect to \(y\).

(ii)     Determine the area of \(R\), giving your answer correct to four significant figures.[3]

b.Find the exact volume generated when the region \(R\) is rotated through \(2\pi \) radians about the \(y\)-axis.[3]

 
▶️Answer/Explanation

Markscheme

(i)     \({\text{area}} = \int_2^4 {\sqrt {y – 2} {\text{d}}y} \)     M1A1

(ii)     \( = 1.886{\text{ (4 sf only)}}\)     A1

Note:     Award M0A0A1 for finding \(1.886\) from \(\int_0^{\sqrt 2 } {4 – f(x){\text{d}}x} \).

Award A1FT for a 4sf answer obtained from an integral involving \(x\).

[3 marks]

a.

\({\text{volume}} = \pi \int_2^4 {(y – 2){\text{d}}y} \)     (M1)

Note:     Award M1 for the correct integral with incorrect limits.

\( = \pi \left[ {\frac{{{y^2}}}{2} – 2y} \right]_2^4\)     (A1)

\( = 2\pi {\text{ (exact only)}}\)     A1

[3 marks]

Total [6 marks]

b.

Examiners report

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Question

The following graph shows the two parts of the curve defined by the equation \({x^2}y = 5 – {y^4}\), and the normal to the curve at the point P(2 , 1).

a.Show that there are exactly two points on the curve where the gradient is zero.[7]

b.Find the equation of the normal to the curve at the point P.[5]

c.The normal at P cuts the curve again at the point Q. Find the \(x\)-coordinate of Q.[3]

d.The shaded region is rotated by 2\(\pi \) about the \(y\)-axis. Find the volume of the solid formed.[7]

▶️Answer/Explanation

Markscheme

differentiating implicitly:       M1

\(2xy + {x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} =  – 4{y^3}\frac{{{\text{d}}y}}{{{\text{d}}x}}\)     A1A1

Note: Award A1 for each side.

if \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) then either \(x = 0\) or \(y = 0\)       M1A1

\(x = 0 \Rightarrow \) two solutions for \(y\left( {y =  \pm \sqrt[4]{5}} \right)\)      R1

\(y = 0\) not possible (as 0 ≠ 5)     R1

hence exactly two points      AG

Note: For a solution that only refers to the graph giving two solutions at  \(x = 0\) and no solutions for \(y = 0\) award R1 only.

[7 marks]

a.

at (2, 1)  \(4 + 4\frac{{{\text{d}}y}}{{{\text{d}}x}} =  – 4\frac{{{\text{d}}y}}{{{\text{d}}x}}\)     M1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} =  – \frac{1}{2}\)     (A1)

gradient of normal is 2       M1

1 = 4 + c       (M1)

equation of normal is \(y = 2x – 3\)     A1

[5 marks]

b.

substituting      (M1)

\({x^2}\left( {2x – 3} \right) = 5 – {\left( {2x – 3} \right)^4}\) or \({\left( {\frac{{y + 3}}{2}} \right)^2}\,y = 5 – {y^4}\)       (A1)

\(x = 0.724\)      A1

[3 marks]

c.

recognition of two volumes      (M1)

volume \(1 = \pi \int_1^{\sqrt[4]{5}} {\frac{{5 – {y^4}}}{y}} {\text{d}}y\left( { = 101\pi  = 3.178 \ldots } \right)\)      M1A1A1

Note: Award M1 for attempt to use \(\pi \int {{x^2}} {\text{d}}y\), A1 for limits, A1 for \({\frac{{5 – {y^4}}}{y}}\) Condone omission of \(\pi \) at this stage.

volume 2

EITHER

\( = \frac{1}{3}\pi  \times {2^2} \times 4\left( { = 16.75 \ldots } \right)\)     (M1)(A1)

OR

\( = \pi \int_{ – 3}^1 {{{\left( {\frac{{y + 3}}{2}} \right)}^2}} {\text{d}}y\left( { = \frac{{16\pi }}{3} = 16.75 \ldots } \right)\)     (M1)(A1)

THEN

total volume = 19.9      A1

[7 marks]

d.

Examiners report

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