IB Mathematics AHL 5.15 Indefinite integrals AA HL Paper 3
Question 1
In this question, you will be investigating the family of functions of the form $f(x) = x^{n}e^{-x}$.
Consider the family of functions $f_{n}(x) = x^{n}e^{-x}$, where $x \geq 0$ and $n \in \mathbb{Z}^{+}$.
When $n = 1$, the function $f_{1}(x) = xe^{-x}$ where $x \geq 0$.
Topic – SL:2.6
(a) Sketch the graph of $y = f_{1}(x)$, stating the coordinates of the local maximum point.
Topic – AHL:5.15
(b) Show that the area of the region bounded by the graph $y = f_{1}(x)$, the x-axis and the line $x = b$, where $b > 0$, is given by $\frac{e^{b}-b-1}{e^{b}}$.
You may assume that the total area, $A_{n}$, of the region between the graph $y = f_{n}(x)$ and the x-axis can be written as $A_{n} = \int_{0}^{\infty}f_{n}(x)dx$ and is given by $\lim_{b \rightarrow \infty} \int_{0}^{b}f_{n}(x)dx$.
Topic – AHL:5.17
(c) (i) Use l’Hôpital’s rule to find $\lim_{b \rightarrow \infty} \frac{e^{b}-b-1}{e^{b}}$. You may assume that the condition for applying l’Hôpital’s rule has been met.
Topic – AHL:5.15
(ii) Hence write down the value of $A_{1}$.
You are given that $A_{2} = 2$ and $A_{3} = 6$.
(d) Use your graphic display calculator, and an appropriate value for the upper limit, to determine the value of
Topic – SL:5.14
(i) $A_{4}$;
Topic – SL:5.14
(ii) $A_{5}$.
Topic – AHL:1.3
(e) Suggest an expression for $A_{n}$ in terms of $n$, where $n \in \mathbb{Z}^{+}$.
Topic – AHL:1.7
(f) Use mathematical induction to prove your conjecture from part (e). You may assume that, for any value of $m$, $\lim_{x \rightarrow \infty} x^{m}e^{-x} = 0$.
▶️Answer/Explanation
Solution: –
(a)
A1 for (1, 0.368) or $\left(1, \frac{1}{e}\right)$ labelled at local maximum (accept correct coordinates written away from the graph)
A1 for graph clearly starting at, or passing through, the origin
A1 for correct domain
A1 for correct shape i.e.: single maximum, and asymptotic behaviour (equation not required) (or point of inflexion)
(b) $\int_{0}^{b}xe^{-x}dx$
Use of integration by parts
\[=[-xe^{-x}]_{0}^{b}+\int_{0}^{b}e^{-x}dx\]
\[=[-xe^{-x}]_{0}^{b}-[e^{-x}]_{0}^{b}\]
attempt to substitute limits
\[=-be^{-b}-e^{-b}+1\]
\[=\frac{e^{b}-b-1}{e^{b}}\]
(c) (i) $\lim_{b\to\infty}\frac{e^{b}-b-1}{e^{b}}=\lim_{b\to\infty}1-\frac{b}{e^{b}}-\frac{1}{e^{b}}$
(ii) $\int_{0}^{\infty}xe^{-x}dx=1$
(d) (i) correct integral
24
(ii) 120
(e) $A_{n}=n!$
(f) \[n=1\]
\[A_{1}=1\] and \(1!=1\)
so true for \(n=1\)
assume true for \(n=k, (A_{k}=\int_{0}^{\infty}x^{k}e^{-s}dx=k!)\)
when \(n=k+1\)
attempt to integrate by parts