IBDP Math AI:Topic:AHL 5.11:Definite and indefinite integration-IB style Questions HL Paper 1

Question

. (a) (i) Expand \((\frac{1}{u} + 1)^2\).
(ii) Find \(\int (\frac{1}{x+2}+1)^2 dx\).
The region bounded by \(y = \frac{1}{(x+2)}+1\), x = 0, x = 2 and the x-axis is rotated through \(2\pi\) about the x-axis to form a solid.
(b) Find the volume of the solid formed. Give your answer in the fomr \(\frac{\pi}{4}(a+b In (c))\), where \(a, b , c\epsilon \mathbb{Z}\).

▶️Answer/Explanation

Ans:

(a) (i) \(\frac{1}{u^2}+\frac{2}{u}+1\)
(ii) \(\int (\frac{1}{(x+2)}+1)^2dx
=\int (\frac{1}{(x+2)^2}+\frac{2}{x+2}+1)dx OR \int (\frac{1}{u^2}+\frac{2}{u})du
=-\frac{1}{(x+2)} + 2 In |x+2| + x(+c)\)

Question

The acceleration, a ms-2 , of a particle moving in a horizontal line at time t seconds, t ≥ 0 , is given by a = – (1+v) where v ms-1 is the particle’s velocity and v > -1.
At t = 0 , the particle is at a fixed origin O and has initial velocity v0 ms-1 .

(a) By solving an appropriate differential equation, show that the particle’s velocity at time t is given by v ( t ) = (1 + v0) e-t – 1 . [6]

(b) Initially at O, the particle moves in the positive direction until it reaches its maximum displacement from O. The particle then returns to O.

Let s metres represent the particle’s displacement from O and smax its maximum displacement from O.
(i) Show that the time T taken for the particle to reach smax satisfies the equation eT = 1 + v0 .
(ii) By solving an appropriate differential equation and using the result from part (b) (i), find an expression for smax in terms of v0 . [7]

Let v (T – k) represent the particle’s velocity k seconds before it reaches smax , where v (T – k) = (1 + v0) e-(T – k) – 1 .

(c) By using the result to part (b) (i), show that v (T – k) = ek – 1 . [2]

Similarly, let v (T + k) represent the particle’s velocity k seconds after it reaches smax .

(d) Deduce a similar expression for v (T + k) in terms of k . [2]

(e) Hence, show that v (T – k) + v (T + k) ≥ 0 . [3]

▶️Answer/Explanation

Ans:

(a) Since $a=\frac{\text{d}v}{\text{d}t}$, and $a=-\left(1+v\right)$, we have
$$\begin{eqnarray}
\frac{\text{d}v}{\text{d}t} = -1-v \nonumber \\
\frac{\text{d}v}{\text{d}t}+v = -1.
\end{eqnarray}$$
Let $\text{I}\left(x\right)=\text{e}^{\int 1 \text{d}t}=\text{e}^t$.<br>
Then, we have
$$\begin{eqnarray}
\frac{\text{d}}{\text{d}t}\left(v\text{e}^t\right) = -\text{e}^t \nonumber \\
v\text{e}^t = -\int\text{e}^t \text{d}t \nonumber \\
v\text{e}^t = -\text{e}^t+c \nonumber \\
v = c\text{e}^{-t}-1.
\end{eqnarray}$$
Since $v=v_0$ when $t=0$, we have $c=1+v_0$, i.e., $v=\left(1+v_0\right)\text{e}^{-t}-1$.<br>
(b)(i) At $s_{\text{max}}$, $t=T$, i.e., we have
$$\begin{eqnarray}
v\left(T\right) = 0 \nonumber \\
\left(1+v_0\right)\text{e}^{-T}-1 = 0 \nonumber \\
\left(1+v_0\right)\text{e}^{-T} = 1 \nonumber \\
1+v_0 = \text{e}^T.
\end{eqnarray}$$
(b)(ii) Since $\frac{\text{d}s}{\text{d}t}=\left(1+v_0\right)\text{e}^{-t}-1$, integrating both sides with respect to $t$, we have
$$\begin{eqnarray}
s &=& \int\left(1+v_0\right)\text{e}^{-t}-1\text{d}t \nonumber \\
&=& -\left(1+v_0\right)\text{e}^{-t}-t+c.
\end{eqnarray}$$
When $t=0$, $s=0$, i.e.,
$$\begin{eqnarray}
0 = -\left(1+v_0\right)+c \nonumber \\
c = \left(1+v_0\right).
\end{eqnarray}$$
Thus, $s=\left(1+v_0\right)-\left(1+v_0\right)\text{e}^{-t}-t$.<br>
Since from (b)(i) we have $1+v_0=\text{e}^T$, $T=\ln\left(1+v_0\right)$, i.e., $s_{\text{max}}=v_0-\ln \left(1+v_0\right)$.<br>
(c) Since $\text{e}^T=1+v_0$, we have $\left(1+v_0\right)\text{e}^{-T}=1$, i.e.,
$$\begin{eqnarray}
v\left(T-k\right) &=& \left(1+v_0\right)\text{e}^{-\left(T-k\right)}-1 \nonumber \\
&=& \left(1+v_0\right)\text{e}^{-T}\text{e}^k-1 \nonumber \\
&=& \text{e}^k-1.
\end{eqnarray}$$
(d) Similarly, as in (c),
$$\begin{eqnarray}
v\left(T+k\right) &=& \left(1+v_0\right)\text{e}^{-\left(T+k\right)}-1 \nonumber \\
&=& \left(1+v_0\right)\text{e}^{-T}\text{e}^{-k}-1 \nonumber \\
&=& \text{e}^{-k}-1.
\end{eqnarray}$$
(e)
$$\begin{eqnarray}
v\left(T-k\right)+v\left(T+k\right) &=& \text{e}^k-1+\text{e}^{-k}-1 \nonumber \\
&=& \text{e}^k+\text{e}^{-k}-2 \nonumber \\
&=& \frac{\text{e}^{2k}-2\text{e}^k+1}{\text{e}^k} \nonumber \\
&=& \frac{\left(\text{e}^k-1\right)^2}{\text{e}^k} \geq 0.
\end{eqnarray}$$

Question

a.Find \(\int {(1 + {{\tan }^2}x){\text{d}}x} \).[2]

b.Find \(\int {{{\sin }^2}x{\text{d}}x} \).[3]

▶️/Explanation

Markscheme

\(\int {(1 + {{\tan }^2}x){\text{d}}x}  = \int {{{\sec }^2}x{\text{d}}x = \tan x( + c)} \)     M1A1

[2 marks]

a.

\(\int {{{\sin }^2}x{\text{d}}x}  = \int {\frac{{1 – \cos 2x}}{2}{\text{d}}x} \)     M1A1

\( = \frac{x}{2} – \frac{{\sin 2x}}{4}( + c)\)     A1

Note:     Allow integration by parts followed by trig identity.

Award M1 for parts, A1 for trig identity, A1 final answer.

[3 marks]

Total [5 marks]

b.

Question

A function \(f\) is defined by \(f(x) = \frac{{3x – 2}}{{2x – 1}},{\text{ }}x \in \mathbb{R},{\text{ }}x \ne \frac{1}{2}\).

a.Find an expression for \({f^{ – 1}}(x)\).[4]

b.Given that \(f(x)\) can be written in the form \(f(x) = A + \frac{B}{{2x – 1}}\), find the values of the constants \(A\) and \(B\).[2]

c.Hence, write down \(\int {\frac{{3x – 2}}{{2x – 1}}} {\text{d}}x\).[1]

▶️Answer/Explanation

Markscheme

\(f:x \to y = \frac{{3x – 2}}{{2x – 1}}\;\;\;{f^{ – 1}}:y \to x\)

\(y = \frac{{3x – 2}}{{2x – 1}} \Rightarrow 3x – 2 = 2xy – y\)     M1

\( \Rightarrow 3x – 2xy =  – y + 2\)     M1

\(x(3 – 2y) = 2 – y\)

\(x = \frac{{2 – y}}{{3 – 2y}}\)     A1

\(\left( {{f^{ – 1}}(y) = \frac{{2 – y}}{{3 – 2y}}} \right)\)

\({f^{ – 1}}(x) = \frac{{2 – x}}{{3 – 2x}}\;\;\;\left( {x \ne \frac{3}{2}} \right)\)     A1

Note:     \(x\) and \(y\) might be interchanged earlier.

Note:     First M1 is for interchange of variables second M1 for manipulation

Note:     Final answer must be a function of \(x\)

[4 marks]

a.

\(\frac{{3x – 2}}{{2x – 1}} = A + \frac{B}{{2x – 1}} \Rightarrow 3x – 2 = A(2x – 1) + B\)

equating coefficients \(3 = 2A\) and \( – 2 =  – A + B\)     (M1)

\(A = \frac{3}{2}\) and \(B =  – \frac{1}{2}\)     A1

Note:     Could also be done by division or substitution of values.

[2 marks]

b.

\(\int {f(x){\text{d}}x = \frac{3}{2}x – \frac{1}{4}\ln \left| {2x – 1} \right| + c} \)     A1

Note:     accept equivalent e.g. \(\ln \left| {4x – 2} \right|\)

[1 mark]

Total [7 marks]

c.

Question

    1. Find \(\int_{0}^{2}\frac{1}{4+x^2}dx\)
    2. Find \(k\) given that \(\int_{0}^{k}\frac{1}{4+x^2}dx=\frac{\pi}{6}\)
▶️/Explanation

Ans:

  1. \(\frac{\pi}{8}\)
  2. \(k=2\sqrt{3}\)
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