# IBDP Maths analysis and approaches Topic: AHL 5.16 Integration by parts HL Paper 2

### Question

A.Prove by mathematical induction that, for $$n \in {\mathbb{Z}^ + }$$,

$1 + 2\left( {\frac{1}{2}} \right) + 3{\left( {\frac{1}{2}} \right)^2} + 4{\left( {\frac{1}{2}} \right)^3} + … + n{\left( {\frac{1}{2}} \right)^{n – 1}} = 4 – \frac{{n + 2}}{{{2^{n – 1}}}}.$[8]

B.(a)     Using integration by parts, show that $$\int {{{\text{e}}^{2x}}\sin x{\text{d}}x = \frac{1}{5}{{\text{e}}^{2x}}} (2\sin x – \cos x) + C$$ .

(b)     Solve the differential equation $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \sqrt {1 – {y^2}} {{\text{e}}^{2x}}\sin x$$, given that y = 0 when x = 0,

writing your answer in the form $$y = f(x)$$ .

(c)     (i)     Sketch the graph of $$y = f(x)$$ , found in part (b), for $$0 \leqslant x \leqslant 1.5$$ .

Determine the coordinates of the point P, the first positive intercept on the x-axis, and mark it on your sketch.

(ii)     The region bounded by the graph of $$y = f(x)$$ and the x-axis, between the origin and P, is rotated 360° about the x-axis to form a solid of revolution.

Calculate the volume of this solid.[17]

### Markscheme

prove that $$1 + 2\left( {\frac{1}{2}} \right) + 3{\left( {\frac{1}{2}} \right)^2} + 4{\left( {\frac{1}{2}} \right)^3} + … + n{\left( {\frac{1}{2}} \right)^{n – 1}} = 4 – \frac{{n + 2}}{{{2^{n – 1}}}}$$

for n = 1

$${\text{LHS}} = 1,{\text{ RHS}} = 4 – \frac{{1 + 2}}{{{2^0}}} = 4 – 3 = 1$$

so true for n = 1     R1

assume true for n = k     M1

so $$1 + 2\left( {\frac{1}{2}} \right) + 3{\left( {\frac{1}{2}} \right)^2} + 4{\left( {\frac{1}{2}} \right)^3} + … + k{\left( {\frac{1}{2}} \right)^{k – 1}} = 4 – \frac{{k + 2}}{{{2^{k – 1}}}}$$

now for n = k +1

LHS: $$1 + 2\left( {\frac{1}{2}} \right) + 3{\left( {\frac{1}{2}} \right)^2} + 4{\left( {\frac{1}{2}} \right)^3} + … + k{\left( {\frac{1}{2}} \right)^{k – 1}} + (k + 1){\left( {\frac{1}{2}} \right)^k}$$     A1

$$= 4 – \frac{{k + 2}}{{{2^{k – 1}}}} + (k + 1){\left( {\frac{1}{2}} \right)^k}$$     M1A1

$$= 4 – \frac{{2(k + 2)}}{{{2^k}}} + \frac{{k + 1}}{{{2^k}}}\,\,\,\,\,$$(or equivalent)     A1

$$= 4 – \frac{{(k + 1) + 2}}{{{2^{(k + 1) – 1}}}}\,\,\,\,\,$$(accept $$4 – \frac{{k + 3}}{{{2^k}}}$$)     A1

Therefore if it is true for n = k it is true for n = k + 1. It has been shown to be true for n = 1 so it is true for all $$n{\text{ }}( \in {\mathbb{Z}^ + })$$.     R1

Note: To obtain the final R mark, a reasonable attempt at induction must have been made.

[8 marks]

A.

(a)

METHOD 1

$$\int {{{\text{e}}^{2x}}\sin x{\text{d}}x = – \cos x{{\text{e}}^{2x}} + \int {2{{\text{e}}^{2x}}\cos x{\text{d}}x} }$$     M1A1A1

$$= – \cos x{{\text{e}}^{2x}} + 2{{\text{e}}^{2x}}\sin x – \int {4{{\text{e}}^{2x}}\sin x{\text{d}}x}$$     A1A1

$$5\int {{{\text{e}}^{2x}}\sin x{\text{d}}x = – \cos x{{\text{e}}^{2x}} + 2{{\text{e}}^{2x}}\sin x}$$     M1

$$\int {{{\text{e}}^{2x}}\sin x{\text{d}}x = \frac{1}{5}{{\text{e}}^{2x}}(2\sin x – \cos x) + C}$$     AG

METHOD 2

$$\int {\sin x{{\text{e}}^{2x}}{\text{d}}x = \frac{{\sin x{{\text{e}}^{2x}}}}{2} – \int {\cos x\frac{{{{\text{e}}^{2x}}}}{2}{\text{d}}x} }$$     M1A1A1

$$= \frac{{\sin x{{\text{e}}^{2x}}}}{2} – \cos x\frac{{{{\text{e}}^{2x}}}}{4} – \int {\sin x\frac{{{{\text{e}}^{2x}}}}{4}{\text{d}}x}$$     A1A1

$$\frac{5}{4}\int {{{\text{e}}^{2x}}\sin x{\text{d}}x = \frac{{{{\text{e}}^{2x}}\sin x}}{2} – \frac{{\cos x{{\text{e}}^{2x}}}}{4}}$$     M1

$$\int {{{\text{e}}^{2x}}\sin x{\text{d}}x = \frac{1}{5}{{\text{e}}^{2x}}(2\sin x – \cos x) + C}$$     AG

[6 marks]

(b)

$$\int {\frac{{{\text{d}}y}}{{\sqrt {1 – {y^2}} }} = \int {{{\text{e}}^{2x}}\sin x{\text{d}}x} }$$     M1A1

$$\arcsin y = \frac{1}{5}{{\text{e}}^{2x}}(2\sin x – \cos x)( + C)$$     A1

when $$x = 0,{\text{ }}y = 0 \Rightarrow C = \frac{1}{5}$$     M1

$$y = \sin \left( {\frac{1}{5}{{\text{e}}^{2x}}(2\sin x – \cos x) + \frac{1}{5}} \right)$$     A1

[5 marks]

(c)

(i)         A1

P is (1.16, 0)     A1

Note: Award A1 for 1.16 seen anywhere, A1 for complete sketch.

Note: Allow FT on their answer from (b)

(ii)     $$V = \int_0^{1.162…} {\pi {y^2}{\text{d}}x}$$     M1A1

$$= 1.05$$     A2

Note: Allow FT on their answers from (b) and (c)(i).

[6 marks]

B.

### Question

a.Find $$\int {x{{\sec }^2}x{\text{d}}x}$$.[4]

b.Determine the value of m if $$\int_0^m {x{{\sec }^2}x{\text{d}}x = 0.5}$$, where m > 0.[2]

### Markscheme

$$\int {x{{\sec }^2}x{\text{d}}x} = x\tan x – \int {1 \times \tan x{\text{d}}x}$$     M1A1

$$= x\tan x + \ln \left| {\cos x} \right|( + c){\text{ }}\left( { = x\tan x – \ln \left| {\sec x} \right|( + c)} \right)$$     M1A1

[4 marks]

a.

attempting to solve an appropriate equation eg $$m\tan m + \ln (\cos m) = 0.5$$     (M1)

m = 0.822     A1

Note: Award A1 if m = 0.822 is specified with other positive solutions.

[2 marks]

b.
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