IB Math Analysis & Approaches Question bank-Topic: SL 5.5 integration as anti-differentiation SL Paper 1

Question

Find  \(\int {\frac{1}{{2x + 3}}} {\rm{d}}x\) .

[2]
a.

Given that \(\int_0^3 {\frac{1}{{2x + 3}}} {\rm{d}}x = \ln \sqrt P \) , find the value of P.

[4]
b.
Answer/Explanation

Markscheme

\(\int {\frac{1}{{2x + 3}}} {\rm{d}}x = \frac{1}{2}\ln (2x + 3) + C\)  (accept \(\frac{1}{2}\ln |(2x + 3)| + C\) )    A1A1     N2

[2 marks]

a.

\(\int_0^3 {\frac{1}{{2x + 3}}} {\rm{d}}x = \left[ {\frac{1}{2}\ln (2x + 3)} \right]_0^3\)

evidence of substitution of limits     (M1)

e.g.\(\frac{1}{2}\ln 9 – \frac{1}{2}\ln 3\)

evidence of correctly using \(\ln a – \ln b = \ln \frac{a}{b}\) (seen anywhere)     (A1)

e.g. \(\frac{1}{2}\ln 3\)

evidence of correctly using \(a\ln b = \ln {b^a}\) (seen anywhere)     (A1)

e.g. \(\ln \sqrt {\frac{9}{3}} \)

\(P = 3\) (accept \(\ln \sqrt 3 \) )     A1     N2

[4 marks]

b.

Question

A function f (x) has derivative f ′(x) = 3x2 + 18x. The graph of f has an x-intercept at x = −1.

Find f (x).

[6]
a.

The graph of f has a point of inflexion at x = p. Find p.

[4]
b.

Find the values of x for which the graph of f is concave-down.

[3]
c.
Answer/Explanation

Markscheme

evidence of integration       (M1)

eg  \(\int {f’\left( x \right)} \)

correct integration (accept absence of C)       (A1)(A1)

eg  \({x^3} + \frac{{18}}{2}{x^2} + C,\,\,{x^3} + 9{x^2}\)

attempt to substitute x = −1 into their = 0 (must have C)      M1

eg  \({\left( { – 1} \right)^3} + 9{\left( { – 1} \right)^2} + C = 0,\,\, – 1 + 9 + C = 0\)

Note: Award M0 if they substitute into original or differentiated function.

correct working       (A1)

eg  \(8 + C = 0,\,\,\,C =  – 8\)

\(f\left( x \right) = {x^3} + 9{x^2} – 8\)      A1 N5

[6 marks]

a.

METHOD 1 (using 2nd derivative)

recognizing that f” = 0 (seen anywhere)      M1

correct expression for f”      (A1)

eg   6x + 18, 6p + 18

correct working      (A1)

6+ 18 = 0

p = −3       A1 N3

METHOD 1 (using 1st derivative)

recognizing the vertex of f′ is needed       (M2)

eg   \( – \frac{b}{{2a}}\) (must be clear this is for f′)

correct substitution      (A1)

eg   \(\frac{{ – 18}}{{2 \times 3}}\)

p = −3       A1 N3

[4 marks]

b.

valid attempt to use f” (x) to determine concavity      (M1)

eg   f” (x) < 0, f” (−2), f” (−4),  6x + 18 ≤ 0 

correct working       (A1)

eg   6x + 18 < 0, f” (−2) = 6, f” (−4) = −6 

f concave down for x < −3 (do not accept ≤ −3)       A1 N2

[3 marks]

c.

Question

Let \(f\left( x \right) = 6{x^2} – 3x\). The graph of \(f\) is shown in the following diagram.

Find \(\int {\left( {6{x^2} – 3x} \right){\text{d}}x} \).

[2]
a.

Find the area of the region enclosed by the graph of \(f\), the x-axis and the lines x = 1 and x = 2 .

[4]
b.
Answer/Explanation

Markscheme

\(2{x^3} – \frac{{3{x^2}}}{2} + c\,\,\,\left( {{\text{accept}}\,\,\frac{{6{x^3}}}{3} – \frac{{3{x^2}}}{2} + c} \right)\)     A1A1 N2

Notes: Award A1A0 for both correct terms if +c is omitted.
Award A1A0 for one correct term eg \(2{x^3} + c\).
Award A1A0 if both terms are correct, but candidate attempts further working to solve for c.

[2 marks]

a.

substitution of limits or function (A1)

eg  \(\int_1^2 {f\left( x \right)} \,{\text{d}}x,\,\,\left[ {2{x^3} – \frac{{3{x^2}}}{2}} \right]_1^2\)

substituting limits into their integrated function and subtracting     (M1)

eg  \(\frac{{6 \times {2^3}}}{3} – \frac{{3 \times {2^2}}}{2} – \left( {\frac{{6 \times {1^3}}}{3} + \frac{{3 \times {1^2}}}{2}} \right)\)

Note: Award M0 if substituted into original function.

correct working      (A1)

eg  \(\frac{{6 \times 8}}{3} – \frac{{3 \times 4}}{2} – \frac{{6 \times 1}}{3} + \frac{{3 \times 1}}{2},\,\,\left( {16 – 6} \right) – \left( {2 – \frac{3}{2}} \right)\)

\(\frac{{19}}{2}\)     A1 N3

[4 marks]

b.

Leave a Reply