IB DP Maths Topic 9.4 Fundamental theorem of calculus HL Paper 3

 

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Question

A function \(f\) is given by \(f(x) = \int_0^x {\ln (2 + \sin t){\text{d}}t} \).

Write down \(f'(x)\).

[1]
a.

By differentiating \(f({x^2})\), obtain an expression for the derivative of \(\int_0^{{x^2}} {\ln (2 + \sin t){\text{d}}t} \) with respect to \(x\).

[3]
b.

Hence obtain an expression for the derivative of \(\int_x^{{x^2}} {\ln (2 + \sin t){\text{d}}t} \) with respect to \(x\).

[3]
c.
Answer/Explanation

Markscheme

\(\ln (2 + \sin x)\)    A1

Note:     Do not accept \(\ln (2 + \sin t)\).

[1 mark]

a.

attempt to use chain rule     (M1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {f({x^2})} \right) = 2xf'({x^2})\)    (A1)

\( = 2x\ln \left( {2 + \sin ({x^2})} \right)\)    A1

[3 marks]

b.

\(\int_x^{{x^2}} {\ln (2 + \sin t){\text{d}}t = \int_0^{{x^2}} {\ln (2 + \sin t){\text{d}}t – \int_0^x {\ln (2 + \sin t){\text{d}}t} } } \)    (M1)(A1)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\int_x^{{x^2}} {\ln (2 + \sin t){\text{d}}t} } \right) = 2x\ln \left( {2 + \sin ({x^2})} \right) – \ln (2 + \sin x)\)    A1

[3 marks]

c.

Examiners report

Many candidates answered this question well. Many others showed no knowledge of this part of the option; candidates that recognized the Fundamental Theorem of Calculus answered this question well. In general the scores were either very low or full marks.

a.

Many candidates answered this question well. Many others showed no knowledge of this part of the option; candidates that recognized the Fundamental Theorem of Calculus answered this question well. In general the scores were either very low or full marks.

b.

Many candidates answered this question well. Many others showed no knowledge of this part of the option; candidates that recognized the Fundamental Theorem of Calculus answered this question well. In general the scores were either very low or full marks.

c.

Question

Find the value of \(\int\limits_4^\infty  {\frac{1}{{{x^3}}}{\text{d}}x} \).

[3]
a.

Illustrate graphically the inequality \(\sum\limits_{n = 5}^\infty  {\frac{1}{{{n^3}}}}  < \int\limits_4^\infty  {\frac{1}{{{x^3}}}{\text{d}}x}  < \sum\limits_{n = 4}^\infty  {\frac{1}{{{n^3}}}} \).

[4]
b.

Hence write down a lower bound for \(\sum\limits_{n = 4}^\infty  {\frac{1}{{{n^3}}}} \).

[1]
c.

Find an upper bound for \(\sum\limits_{n = 4}^\infty  {\frac{1}{{{n^3}}}} \).

[3]
d.
Answer/Explanation

Markscheme

\(\int\limits_4^\infty  {\frac{1}{{{x^3}}}{\text{d}}x}  = \mathop {{\text{lim}}}\limits_{R \to \infty } \int\limits_4^R {\frac{1}{{{x^3}}}{\text{d}}x} \)      (A1)

Note: The above A1 for using a limit can be awarded at any stage.
Condone the use of \(\mathop {{\text{lim}}}\limits_{x \to \infty } \).

Do not award this mark to candidates who use \(\infty \) as the upper limit throughout.

= \(\mathop {{\text{lim}}}\limits_{R \to \infty } \left[ { – \frac{1}{2}{x^{ – 2}}} \right]_4^R\left( { = \left[ { – \frac{1}{2}{x^{ – 2}}} \right]_4^\infty } \right)\)     M1

\( = \mathop {{\text{lim}}}\limits_{R \to \infty } \left( { – \frac{1}{2}\left( {{R^{ – 2}} – {4^{ – 2}}} \right)} \right)\)

\( = \frac{1}{{32}}\)     A1

[3 marks]

a.

     A1A1A1A1

A1 for the curve
A1 for rectangles starting at \(x = 4\)
A1 for at least three upper rectangles
A1 for at least three lower rectangles

Note: Award A0A1 for two upper rectangles and two lower rectangles.

sum of areas of the lower rectangles < the area under the curve < the sum of the areas of the upper rectangles so

\(\sum\limits_{n = 5}^\infty  {\frac{1}{{{n^3}}}}  < \int\limits_4^\infty  {\frac{1}{{{x^3}}}{\text{d}}x}  < \sum\limits_{n = 4}^\infty  {\frac{1}{{{n^3}}}} \)      AG

[4 marks]

b.

a lower bound is \(\frac{1}{{32}}\)     A1

Note: Allow FT from part (a).

[1 mark]

c.

METHOD 1

\(\sum\limits_{n = 5}^\infty  {\frac{1}{{{n^3}}}}  < \frac{1}{{32}}\)    (M1)

\(\frac{1}{{64}} + \sum\limits_{n = 5}^\infty  {\frac{1}{{{n^3}}}}  = \frac{1}{{32}} + \frac{1}{{64}}\)     (M1)

\(\sum\limits_{n = 4}^\infty  {\frac{1}{{{n^3}}}}  < \frac{3}{{64}}\), an upper bound      A1

Note: Allow FT from part (a).

METHOD 2

changing the lower limit in the inequality in part (b) gives

\(\sum\limits_{n = 4}^\infty  {\frac{1}{{{n^3}}}}  < \int\limits_3^\infty  {\frac{1}{{{x^3}}}{\text{d}}x} \left( { < \sum\limits_{n = 3}^\infty  {\frac{1}{{{n^3}}}} } \right)\)     (A1)

\(\sum\limits_{n = 4}^\infty  {\frac{1}{{{n^3}}}}  < \mathop {{\text{lim}}}\limits_{R \to \infty } \left[ { – \frac{1}{2}{x^{ – 2}}} \right]_3^R\)     (M1)

\(\sum\limits_{n = 4}^\infty  {\frac{1}{{{n^3}}}}  < \frac{1}{{18}}\), an upper bound     A1

Note: Condone candidates who do not use a limit.

[3 marks]

d.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

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