🔬 Tutorial problems kappa

Note

This problems are designed to help you practice the concepts covered in the lectures. Not all problems may be covered in the tutorial, those left are for additional practice. The symbol 🍹 indicates additional problems.

\(\kappa\).1

Formulation

In the manufacture of a product, the marginal cost of producing \(x\) units is \(C'(x)\) and fixed costs are \(C(0)\). Find the total cost function \(C(x)\) when

(a) \(C'(x) = 3x + 4\) and \(C(0) = 40\)

(b) \(C'(x) = ax + b\) and \(C(0) = C_0\)

Source: SHSC Exercise 9.1.2, page 324

Solution

(a)

\(C'(x) = 3x + 4\) and \(C(0) = 40\)

\[\begin{split} \begin{array}{ll} C(x) &= \int (3x + 4) \mathop{dx} \\ &= \frac{3}{2}x^2 + 4x + C \end{array} \end{split}\]

\(C(0) = 40 \iff \frac{3}{2}(0)^2 + 4(0) + C = 40 \iff C = 40\)

Therefore \(C(x) = \frac{3}{2}x^2 + 4x + 40\).

(b)

\(C'(x) = ax + b\) and \(C(0) = C_0\)

\[\begin{split} \begin{array}{ll} C(x) &= \int (ax + b) \mathop{dx} \\ &= \frac{a}{2}x^2 + bx + C \end{array} \end{split}\]

\(C(0) = C_0 \iff \frac{a}{2}(0)^2 + b(0) + C = C_0 \iff C = C_0\)

Therefore \(C(x) = \frac{a}{2}x^2 + bx + C_0\).

Source: SHSC Exercise 9.1.2, page 324

\(\kappa\).2

Formulation

Find the following integrals:

(a) \(\displaystyle \int \frac{(y - 2)^2}{\sqrt{y}} \mathop{dy}\)

(b) \(\displaystyle \int \frac{x^3}{x + 1} \mathop{dx}\)

(c) \(\displaystyle \int x(1 + x^2)^{15} \mathop{dx}\)

Hints:

• In part (a), first expand \((y - 2)^2\), and then divide each term by \(\sqrt{y}\)

• In part (b), note that \(x^3 = x^2 - x +1 -\frac{1}{x+1}\) (this is polynomial division as in Section 4.7 of SHSC)

• In part (c), what is the derivative of \((1 + x^2)^{16}\)?

Source: SHSC Exercise 9.1.4, page 324

Solution

(a)

\[\begin{split} \begin{array}{ll} \int \frac{(y - 2)^2}{\sqrt{y}} \mathop{dy} &= \int \frac{y^2 - 4y + 4}{\sqrt{y}} \mathop{dy} \\ &= \int y^{1.5} - 4y^{0.5} + 4y^{-0.5} \mathop{dy} \\ &= \frac{1}{2.5}y^{2.5} - 4(\frac{1}{1.5})y^{1.5} + 4(\frac{1}{0.5})y^{0.5} + C \\ &= \frac{2}{5}y^{2.5} - 4(\frac{2}{3})y^{1.5} + 4(2)y^{0.5} + C \\ &= \frac{2}{5}y^{2.5} - \frac{8}{3}y^{1.5} + 8y^{0.5} + C \\ &= \frac{2}{5}y^2 \sqrt{y} - \frac{8}{3}y \sqrt{y} + 8 \sqrt{y} + C \end{array} \end{split}\]

(b)

\[\begin{split} \begin{array}{ll} \int \frac{x^3}{x + 1} \mathop{dx} &= \int x^2 - x + 1 - \frac{1}{x + 1} \mathop{dx} \\ &= \frac{1}{3}x^3 - \frac{1}{2}x^2 + x - \ln |x + 1| + C \end{array} \end{split}\]

(c)

First notice that

\[ \frac{d}{dx}[(1 + x^2)^{16}] = (16)(1 + x^2)^{15}(2x) = 32x(1 + x^2)^{15} \]

Thus we have

\[ \int x(1 + x^2)^{15} \mathop{dx} = \frac{1}{32} \int 32 x(1 + x^2)^{15} \mathop{dx} = \frac{1}{32}(1 + x^2)^{16} + C \]

Source: SHSC Exercise 9.1.4, page 324

\(\kappa\).3

Formulation

Find the general form of a function \(f\) whose second derivative is \(x^2\).

If we require in addition that \(f(0) = 1\) and \(f'(0) = -1\), what is \(f(x)\)?

Source: SHSC Exercise 9.1.10, page 325

Solution

Given: \(f''(x) = x^2\).

Then \(f'(x) = \int x^2 \mathop{dx} = \frac{1}{3} x^3 + A\), where \(A\) is an arbitrary constant.

Then \(f(x) = \int \frac{1}{3} x^3 + A \mathop{dx} = \frac{1}{12} x^4 + Ax + B\), where \(A\) and \(B\) are arbitrary constants. This is the general form of the function whose second derivative is \(x^2\).

We also know \(f'(0) = -1 \iff \frac{1}{3} (0) + A = -1 \iff A = -1\), and \(f(0) = 1 \iff B = 1\) by the same logic.

Therefore in this case \(f(x) = \frac{1}{12} x^4 - x + 1\).

Source: SHSC Exercise 9.1.10, page 325

\(\kappa\).4

Formulation

Find the following integrals:

(a) \(\displaystyle \int_{1}^{\infty} \frac{5}{x^5} \mathop{dx}\)

(b) \(\displaystyle \int_{0}^{1} x^3(1 + x^4)^4 \mathop{dx}\)

(c) \(\displaystyle \int_{0}^{\infty} \frac{-5t}{e^t} \mathop{dt}\)

(d) \(\displaystyle \int_{1}^{e} (\ln x)^2 \mathop{dx}\)

(e) \(\displaystyle \int_{0}^{2} x^2 \sqrt{x^3 + 1} \mathop{dx}\)

(f) \(\displaystyle \int_{-\infty}^{0} \frac{e^{3z}}{e^{3z} + 5} \mathop{dz}\)

(g) \(\displaystyle \int_{1/2}^{e/2} x^3 (\ln 2x) \mathop{dx}\)

(h) \(\displaystyle \int_{1}^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \mathop{dx}\)

Source: SHSC Ch 9 Review Exercise 4, page 372

Solution

(a)

\[\begin{split} \begin{array}{ll} \int_{1}^{\infty} \frac{5}{x^5} \mathop{dx} &= 5 \int_{1}^{\infty} x^{-5} \mathop{dx} \\ &= 5 \lim_{a \to \infty} \int_{1}^{a} x^{-5} \mathop{dx} \\ &= 5 \lim_{a \to \infty} (\frac{-4}{x^4})\Big|_{1}^{a} \\ &= 5 [0 - \frac{-1}{4}(1)^{-4}] \\ &= 5/4 \end{array} \end{split}\]

(b)

Let \(u = (x^4 + 1)\). Then \(du = 4x^3 dx\). Further, \(x = 0 \Rightarrow u = 1\) and \(x = 1 \Rightarrow u = 2\).

\[\begin{split} \begin{array}{ll} \int_{0}^{1} x^3(1 + x^4)^4 \mathop{dx} &= \frac{1}{4} \int_{1}^{2} u^4 \mathop{du} \\ &= \frac{1}{4} \frac{1}{5} \Big|_{1}^{2} (u^5) \\ &= \frac{1}{20} (32 - 1) \\ &= 31/20 \end{array} \end{split}\]

(c)

\[\begin{split} \begin{array}{ll} \int_{0}^{\infty} \frac{-5t}{e^t} \mathop{dt} &= -5 \int_{0}^{\infty} t e^{-t} \mathop{dt} \\ &= -5 [\Big|_{0}^{\infty} t (-e^{-t}) - \int_{0}^{\infty} (1)(-e^{-t}) \mathop{dt}] \\ &= -5 [0 + \int_{0}^{\infty} e^{-t} \mathop{dt}] \\ &= -5 [\Big|_{0}^{\infty} (-e^{-t}) \mathop{dt}] \\ &= -5 [- \Big|_{0}^{\infty} (e^{-t}) \mathop{dt}] \\ &= -5 [- (0 - 1)] \mathop{dt}] \\ &= -5 \end{array} \end{split}\]

(d)

Let \(f(x) = (\ln x)^2\) and let \(g'(x) = 1\). Then \(f'(x) = 2 \ln x \frac{1}{x}\) and \(g(x) = x\).

\[\begin{split} \begin{array}{ll} \int_{1}^{e} (\ln x)^2 \mathop{dx} &= \Big|_{1}^{e} [(\ln x)^2 x] - \int_{1}^{e} 2 \ln x \frac{1}{x} x \mathop{dx} \\ &= \Big|_{1}^{e} [x (\ln x)^2] - 2 \int_{1}^{e} \ln x \mathop{dx} \\ &= [e (\ln e)^2 - 1 (\ln 1)^2] - 2 \int_{1}^{e} \ln x \mathop{dx} \\ &= e - 2 \int_{1}^{e} \ln x \mathop{dx} \\ &= e - 2 \int_{1}^{e} \ln x \mathop{dx} \\ &= e - 2 \end{array} \end{split}\]

noting that if we let \(h(x) = \ln x\) and \(j'(x) = 1\), then \(h'(x) = \frac{1}{x}\) and \(j(x) = x\), so

\[ \int_{1}^{e} \ln x \mathop{dx} = \Big|_{1}^{e} [x \ln x] - \int_{1}^{e} \frac{1}{x} x \mathop{dx} = (e - 0) - (e - 1) = 1 \]

(e)

Let \(u = x^3 + 1 \Rightarrow \mathop{du} = 3x^2 \mathop {dx}\). When \(x = 0 \Rightarrow u = 1\) and \(x = 2 \Rightarrow u = 9\).

\[\begin{split} \begin{array}{ll} \int_{0}^{2} x^2 \sqrt{x^3 + 1} \mathop{dx} &= \frac{1}{3} \int_{1}^{9} \sqrt{u} \mathop{du} \\ &= \frac{1}{3} \frac{2}{3} \Big|_{1}^{9} u^{1.5} \\ &= \frac{2}{9} (9^{1.5} - 1^{1.5}) \\ &= \frac{2}{9} (27 - 1) \\ &= 52/9 \end{array} \end{split}\]

(f)

Let \(u = e^{3z} \Rightarrow \mathop{du} = 3 e^{3z} \mathop{dz}\). When \(z \rightarrow -\infty \Rightarrow u \rightarrow 0\), and \(z = 0 \Rightarrow u = 1\).

\[\begin{split} \begin{array}{ll} \int_{-\infty}^{0} \frac{e^{3z}}{e^{3z} + 5} \mathop{dz} &= \frac{1}{3} \int_{0}^{1} \frac{1}{u + 5} \mathop{du} \\ &= \frac{1}{3} \Big|_{0}^{1} \ln |u + 5| \mathop{du} \\ &= \frac{1}{3} [\ln |1 + 5| - \ln |0 + 5|] \\ &= \frac{1}{3} [\ln 6 - \ln 5] \\ &= \frac{1}{3} \ln(\frac{6}{5}) \end{array} \end{split}\]

(g)

Let \(f(x) = \ln 2x\) and let \(g'(x) = x^3\). Then \(f'(x) = \frac{1}{x}\) and \(g(x) = \frac{1}{4} x^4\).

\[\begin{split} \begin{array}{ll} \int_{1/2}^{e/2} x^3 (\ln (2x)) \mathop{dx} &= \frac{1}{4} \Big|_{1/2}^{e/2} [\ln (2x) x^4] - \frac{1}{4} \int_{1/2}^{e/2} \frac{1}{x} x^4 \mathop{dx} \\ &= \frac{1}{4} \Big|_{1/2}^{e/2} [\ln (2x) x^4] - \frac{1}{4} \int_{1/2}^{e/2} x^3 \mathop{dx} \\ &= \frac{1}{4} \Big|_{1/2}^{e/2} [\ln (2x) x^4] - \frac{1}{16} \Big|_{1/2}^{e/2} x^4 \\ &= \frac{1}{4} [\frac{e^4}{2^4} - 0] - \frac{1}{16} [\frac{e^4}{2^4} - \frac{1}{2^4}] \\ &= \frac{e^4}{2^6} - \frac{e^4 - 1}{2^8} \\ &= \frac{4e^4}{2^8} - \frac{e^4 - 1}{2^8} \\ &= (3e^4 + 1)/256 \end{array} \end{split}\]

(h)

Let \(u = \sqrt{x}\). Then \(du = \frac{1}{2} x^-0.5 \mathop{dx}\). When \(x = 1 \Rightarrow u = 1\) and \(x \rightarrow \infty \Rightarrow u \rightarrow \infty\).

\[\begin{split} \begin{array}{ll} \int_{1}^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \mathop{dx} &= 2 \int_{1}^{\infty} e^{-u} \mathop{du} \\ &= -2 \Big|_{1}^{\infty} e^{-u} \\ &= -2 [0 - e^{-1}] \\ &= 2/e \end{array} \end{split}\]

Source: SHSC Review Exercise 4, page 372

\(\kappa\).5

Formulation

Compute the area bounded by the graph of \(f(x) = \frac{1}{2}(e^x + e^{-x})\), the \(x\)-axis, and the two lines \(x = -1\) and \(x = 1\).

Source: SHSC Exercise 9.2.4, page 331

Solution

\[\begin{split} \begin{array}{ll} \int_{-1}^{1} \frac{1}{2}(e^x + e^{-x}) \mathop{dx} &= \frac{1}{2} \int_{-1}^{1} (e^x + e^{-x}) \mathop{dx} \\ &= \frac{1}{2} \Big|_{-1}^{1} (e^x - e^{-x}) \\ &= \frac{1}{2} [(e^1 - e^{(-1)}) - (e^{-1} - e^{-(-1)})] \\ &= \frac{1}{2} [e - e^{-1} - e^{-1} + e)] \\ &= \frac{1}{2} [2e - 2e^{-1}] \\ &= e - e^{-1} \\ \end{array} \end{split}\]

Source: SHSC Exercise 9.2.4, page 331

\(\kappa\).6

Formulation

Example 9.7.1 in SHSC tells us that the exponential distribution in statistics is defined by the density function \(f(x) = \lambda e^{-\lambda x}\), where \(x \geq 0\) and \(\lambda\) is a positive constant.

We now wish to find the following:

(a) \(\displaystyle \int_{0}^{\infty} x \lambda e^{-\lambda x} \mathop{dx}\)

(b) \(\displaystyle \int_{0}^{\infty} (x - 1/\lambda)^2 \lambda e^{-\lambda x} \mathop{dx}\)

(c) \(\displaystyle \int_{0}^{\infty} (x - 1/\lambda)^3 \lambda e^{-\lambda x} \mathop{dx}\)

The three numbers you obtain are called respectively the expectation, the variance, and the third central moment of the exponential distribution.

Source: SHSC Exercise 9.7.3, page 358

Solution

(a)

Use integration by parts.

Let \(g(x) = x\) and \(h'(x) = \lambda e^{-\lambda x}\). Then \(g'(x) = 1\) and \(h(x) = -e^{-\lambda x}\).

Then

\[\begin{split} \begin{array}{ll} \int_{0}^{\infty} x \lambda e^{-\lambda x} \mathop{dx} &= - \Big|_{0}^{\infty} x e^{-\lambda x} - \int_{0}^{\infty} (1) (-e^{-\lambda x}) \mathop{dx} \\ &= - \Big|_{0}^{\infty} x e^{-\lambda x} + \int_{0}^{\infty} e^{-\lambda x} \mathop{dx} \\ &= - \lim\limits_{b \rightarrow \infty} [b e^{-\lambda b} - 0 e^{-\lambda 0}] + \int_{0}^{\infty} e^{-\lambda x} \mathop{dx} \\ &= - \lim\limits_{b \rightarrow \infty} [b e^{-\lambda b}] + \int_{0}^{\infty} e^{-\lambda x} \mathop{dx} \\ &= 0 + \int_{0}^{\infty} e^{-\lambda x} \mathop{dx} \\ &= \lim\limits_{b \rightarrow \infty} [\frac{-1}{\lambda} e^{-\lambda b} - \frac{-1}{\lambda} e^{-\lambda 0}] \\ &= \frac{1}{\lambda} \end{array} \end{split}\]

(b)

Use integration by parts.

Let \(g(x) = (x - \frac{1}{\lambda})^2\) and \(h'(x) = \lambda e^{-\lambda x}\). Then \(g'(x) = 2x - \frac{2}{\lambda}\) and \(h(x) = -e^{-\lambda x}\).

Then

\[\begin{split} \begin{array}{ll} \int_{0}^{\infty} (x - \frac{1}{\lambda})^2 \lambda e^{-\lambda x} \mathop{dx} &= - \Big|_{0}^{\infty} (x - \frac{1}{\lambda})^2 e^{-\lambda x} - \int_{0}^{\infty} (2x - \frac{2}{\lambda}) (-e^{-\lambda x}) \mathop{dx} \\ &= - \lim\limits_{b \rightarrow \infty} [(b - \frac{1}{\lambda})^2 e^{-\lambda b} - (0 - \frac{1}{\lambda})^2 e^{-\lambda 0}] - \int_{0}^{\infty} (2x - \frac{2}{\lambda}) (-e^{-\lambda x}) \mathop{dx} \\ &= - [0 - (-\frac{1}{\lambda})^2 (1)] - \int_{0}^{\infty} (2x - \frac{2}{\lambda}) (-e^{-\lambda x}) \mathop{dx} \\ &= \frac{1}{\lambda^2} - \int_{0}^{\infty} (2x - \frac{2}{\lambda}) (-e^{-\lambda x}) \mathop{dx} \\ &= \frac{1}{\lambda^2} + \int_{0}^{\infty} (2x - \frac{2}{\lambda}) (e^{-\lambda x}) \mathop{dx} \\ &= \frac{1}{\lambda^2} + \int_{0}^{\infty} 2x (e^{-\lambda x}) \mathop{dx} - \int_{0}^{\infty} \frac{2}{\lambda} (e^{-\lambda x}) \mathop{dx} \\ &= \frac{1}{\lambda^2} + \frac{2}{\lambda} \int_{0}^{\infty} x \lambda (e^{-\lambda x}) \mathop{dx} - \frac{2}{\lambda} \int_{0}^{\infty} (e^{-\lambda x}) \mathop{dx} \\ &= \frac{1}{\lambda^2} + \frac{2}{\lambda^2} - \frac{2}{\lambda^2} \\ &= \frac{1}{\lambda^2} \end{array} \end{split}\]

where the second-last equality follows from two applications of our result in part (a).

(c)

Use integration by parts.

Let \(g(x) = (x - \frac{1}{\lambda})^3\) and \(h'(x) = \lambda e^{-\lambda x}\). Then \(g'(x) = 3x^2 - \frac{6}{\lambda}x + \frac{3}{\lambda^2}\) and \(h(x) = -e^{-\lambda x}\).

Then

\[\begin{split} \begin{array}{ll} \int_{0}^{\infty} (x - \frac{1}{\lambda})^3 \lambda e^{-\lambda x} \mathop{dx} &= - \Big|_{0}^{\infty} (x - \frac{1}{\lambda})^3 e^{-\lambda x} - \int_{0}^{\infty} (3x^2 - \frac{6}{\lambda}x + \frac{3}{\lambda^2}) (-e^{-\lambda x}) \mathop{dx} \\ &= - \lim\limits_{b \rightarrow \infty} [(b - \frac{1}{\lambda})^3 e^{-\lambda b} - (0 - \frac{1}{\lambda})^3 e^{-\lambda 0}] - \int_{0}^{\infty} (3x^2 - \frac{6}{\lambda}x + \frac{3}{\lambda^2}) (-e^{-\lambda x}) \mathop{dx} \\ &= -[0 - (-\frac{1}{\lambda})^3 (1)] - \int_{0}^{\infty} (3x^2 - \frac{6}{\lambda}x + \frac{3}{\lambda^2}) (-e^{-\lambda x}) \mathop{dx} \\ &= -\frac{1}{\lambda^3} - \int_{0}^{\infty} (3x^2 - \frac{6}{\lambda}x + \frac{3}{\lambda^2}) (-e^{-\lambda x}) \mathop{dx} \\ &= -\frac{1}{\lambda^3} + \int_{0}^{\infty} (3x^2 - \frac{6}{\lambda}x + \frac{3}{\lambda^2}) (e^{-\lambda x}) \mathop{dx} \\ &= -\frac{1}{\lambda^3} + \int_{0}^{\infty} 3x^2 (e^{-\lambda x}) \mathop{dx} - \int_{0}^{\infty} \frac{6}{\lambda}x (e^{-\lambda x}) \mathop{dx} + \int_{0}^{\infty} \frac{3}{\lambda^2} (e^{-\lambda x}) \mathop{dx} \\ &= -\frac{1}{\lambda^3} + \int_{0}^{\infty} 3x^2 (e^{-\lambda x}) \mathop{dx} - \frac{6}{\lambda^2} \int_{0}^{\infty} x \lambda (e^{-\lambda x}) \mathop{dx} + \frac{3}{\lambda^2} \int_{0}^{\infty} (e^{-\lambda x}) \mathop{dx} \\ &= -\frac{1}{\lambda^3} + \int_{0}^{\infty} 3x^2 (e^{-\lambda x}) \mathop{dx} - \frac{6}{\lambda^3} + \frac{3}{\lambda^3} \\ &= -\frac{1}{\lambda^3} + \int_{0}^{\infty} 3x^2 (e^{-\lambda x}) \mathop{dx} - \frac{3}{\lambda^3} \\ &= -\frac{1}{\lambda^3} + \frac{6}{\lambda^3} - \frac{3}{\lambda^3} \\ &= \frac{2}{\lambda^3} \end{array} \end{split}\]

noting that the second-last equality follows as

\[\begin{split} \begin{array}{ll} \int_{0}^{\infty} 3x^2 (e^{-\lambda x}) \mathop{dx} &= \Big|_{0}^{\infty} 3x^2 (e^{-\lambda x}) - \frac{-1}{\lambda}\int_{0}^{\infty} 6x (e^{-\lambda x}) \mathop{dx} \\ &= 0 + \frac{1}{\lambda} \int_{0}^{\infty} 6x (e^{-\lambda x}) \mathop{dx} \\ &= \frac{6}{\lambda^2} \int_{0}^{\infty} x \lambda (e^{-\lambda x}) \mathop{dx} \\ &= \frac{6}{\lambda^3} \\ \end{array} \end{split}\]

Therefore, for the exponential distribution: the expectation is \(\frac{1}{\lambda}\), the variance is \(\frac{1}{\lambda^2}\) and the third central moment is \(\frac{2}{\lambda^3}\).

Source: SHSC Exercise 9.7.3, page 358