Mathematics for Management -- Supplementary Electronic Materials

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Worked-Out Exercises: Integration by Parts

The following set of exercises is given together with hints, solutions, and solution paths. They are designed such that you can first try to solve them and in case you need help, please, open the "hints" section. For checking your answer, please, open the "solution" section and for getting more details or checking your way of solving the exercise, please, open the "solution path" section.


Exercise 1: Evaluating an Integral

Evaluate the following integral using integration by parts by using the choice $ u = \ln(x) $ and $ \textrm{d} v = \sqrt{x} \, \textrm{d} x $: $$ \int \, \sqrt{x} \cdot \ln(x) \, \textrm{d} x $$

Hint (please, click on the "+" sign to read more) use integration by parts with suitably chosen functions $ u $ and $v $: $$ \int \, u \, \textrm{d} v \, \, = \, \, u \, v \, - \, \int \, v \, \textrm{d} u $$

Solution (please, click on the "+" sign to read more) $ \tfrac{2}{3} x^{3/2} \left( \ln(x) - \tfrac{2}{3} \right) + C $

Solution Path (please, click on the "+" sign to read more) By choosing $$ \begin{array}{r c l c r c l} u & = & \ln(x)\qquad & \qquad & \textrm{d} v & = & \sqrt{x} \, \textrm{d} x \qquad \\[4mm] \textrm{d} u & = & \frac{1}{x} \, \textrm{d} x \qquad & & v & = & \tfrac{2}{3} \, x^{3/2} \qquad \end{array} $$ we obtain $ \int \, \sqrt{x} \cdot \ln(x) \, \textrm{d} x = $ $ \tfrac{2}{3} x^{3/2} \ln(x) - \tfrac{2}{3} \int \, x^{(3/2) - 1} \, \textrm{d} x \, \, = \, \, $ $ \tfrac{2}{3} x^{3/2} \ln(x) - \tfrac{2}{3} \int \, \sqrt{x} \, \textrm{d} x\\ = $ $ =\tfrac{2}{3} x^{3/2} \ln(x) - \tfrac{2}{3} \cdot \left( \tfrac{2}{3} x^{3/2} \right) + C \, \, = \, \, $ $ \tfrac{2}{3} x^{3/2} \left( \ln(x) - \tfrac{2}{3} \right) + C $

Exercise 2: Use Integration by Parts to Prove a Reduction Formula

$$ \int \, \left( \ln(x) \right)^n \, \textrm{d} x \, \, = \, \, x \cdot \left( \ln(x) \right)^n \, - \, n \, \int \, \left( \ln(x) \right)^{n-1} \, \textrm{d} x $$

Hint (please, click on the "+" sign to read more) use integration by parts with suitably chosen functions $ u $ and $v $: $$ \int \, u \, \textrm{d} v \, \, = \, \, u \, v \, - \, \int \, v \, \textrm{d} u $$

Solution Path (please, click on the "+" sign to read more) By choosing $$ \begin{array}{r c l c r c l} u & = & (\ln(x))^n \qquad & \qquad \qquad & \textrm{d} v & = & 1 \, \textrm{d} x \qquad \\[4mm] \textrm{d} u & = & \frac{n}{x} \, (\ln(x))^{n-1} \, \textrm{d} x \qquad & & v & = & x \qquad \end{array} $$ we obtain $$ \int \, \left( \ln(x) \right)^n \, \textrm{d} x = x \cdot \left( \ln(x) \right)^n \, - \, n \, \int \, \left( \ln(x) \right)^{n-1} \cdot \frac{x}{x} \, \textrm{d} x $$

Exercise 3: Applying Integration by Parts Twice

If $ f(0) = g(0) = 0 $ and $ f'' $ and $ g'' $ are continuous, show that $$ \int^a_0 \, f(x) g''(x) \, \textrm{d} x \, \, = \, \, f(a) g'(a) \, - \, f'(a) g(a) \, + \, \int^a_0 \, f''(x) g(x) \, \textrm{d} x $$

Hint (please, click on the "+" sign to read more) use integration by parts several times

Solution Path (please, click on the "+" sign to read more) $$ \int^a_0 \, f(x) g''(x) \, \textrm{d} x \stackrel{\text{(IP)}}{=} f(a) \cdot g'(a) \, - \, \underbrace{f(0) \cdot g'(0)}_{= \, 0} \, - \, \int^a_0 \, f'(x) g'(x) \, \textrm{d} x \ {=}$$ $$\stackrel{\text{(IP)}}{=} f(a) \cdot g'(a) \, - \, f'(a) \cdot g(a) \, + \, \underbrace{f'(0) \cdot g(0)}_{= \, 0} + \, \int^a_0 \, f''(x) g(x) \, \textrm{d} x $$

Exercise 4: Evaluating Integrals

Evaluate the following integrals using integration by parts: $$ {\bf{a)}} {\displaystyle{ \int^5_1 \, \frac{x}{{\rm{e}}^x} \, \textrm{d} x }} $$ as well as $$ {\bf{b)}} {\displaystyle{ \int \, (1+x^2) \cdot {\rm{e}}^{3x} \, \textrm{d} x }} $$

Hint (please, click on the "+" sign to read more) for part $ \bf{(b)} $ and applying integration by parts twice may be necessary.

Solution (please, click on the "+" sign to read more) $$ {\bf{a)}} -6 {\rm{e}}^{-5} + 2 {\rm{e}}^{-1} \, \, \approx \, \, 0.6953312004 $$ $$ {\bf{b)}} \tfrac{1}{3} \, \left( \tfrac{11}{9} - \tfrac{2}{3} x + x^2 \right) \, {\rm{e}}^{3x} $$

Solution Path (please, click on the "+" sign to read more) $$ \int \, u \, \textrm{d} v \, \, = \, \, u \, v \, - \, \int \, v \, \textrm{d} u $$ $ {\bf{a)}} $ by choosing $$ \begin{array}{r c l c r c l} u & = & x \qquad & \qquad & \textrm{d} v & = & {\rm{e}}^{-x} \, \textrm{d} x qquad \\[4mm] \textrm{d} u & = & 1 \, \textrm{d} x \qquad & & v & = & -{\rm{e}}^{-x} \qquad \end{array} $$ we obtain $$ \int^5_1 \, x \cdot {\rm{e}}^{-x} \, \textrm{d} x \big[ -x \cdot {\rm{e}}^{-x} \big]^5_1 \, + \, \int^5_1 \, {\rm{e}}^{-x} \, \textrm{d} x \ {=} $$ $$ \, \, = \, \, \big[ -{\rm{e}}^{-x} \left( x + 1 \right) \big]^5_1 \\[2mm] -6 {\rm{e}}^{-5} + 2 {\rm{e}}^{-1} \, \, \approx \, \, 0.6953312004 $$ $ {\bf{b)}} $ by choosing $$ \begin{array}{r c l c r c l} u & = & (1+x^2) \qquad & \qquad & \textrm{d} v & = & {\rm{e}}^{3x} \, \textrm{d} x \qquad \\[4mm] \textrm{d} u & = & 2x \, \textrm{d} x \qquad & & v & = & \tfrac{1}{3} {\rm{e}}^{3x} \qquad \end{array} $$ we obtain $$ {\bf{b)}} \int \, (1+x^2) \cdot {\rm{e}}^{3x} \, \textrm{d} x \ {=} \tfrac{1}{3} \, (1+x^2) \, {\rm{e}}^{3x} \, - \, \tfrac{2}{3} \int \, x \cdot {\rm{e}}^{3x} \, \textrm{d} x \\ \ {=} $$ $$ \ {=} \tfrac{1}{3} \, (1+x^2) \, {\rm{e}}^{3x} \, - \, \ \tfrac{2}{3} \left( \tfrac{1}{3} \, x \, {\rm{e}}^{3x} \, - \, \tfrac{1}{3} \int \, {\rm{e}}^{3x} \, \textrm{d} x \right) \\ \ {=} $$ $$ \ {=} \tfrac{1}{3} \, \left( \tfrac{11}{9} - \tfrac{2}{3} x + x^2 \right) \, {\rm{e}}^{3x} $$

Exercise 5: Evaluating a Integral

$$ {\displaystyle{ \int^1_0 \, \frac{x^3}{\sqrt{4+x^2}} \, \textrm{d} x }} $$

Hint (please, click on the "+" sign to read more) use integration by parts

Solution (please, click on the "+" sign to read more) $$ \tfrac{16}{3} - \tfrac{7}{3} \sqrt{5} $$

Solution Path (please, click on the "+" sign to read more) by choosing $$ \begin{array}{r c l c r c l} u & = & x^2 \qquad & \qquad \qquad & \textrm{d} v & = & \frac{x}{\sqrt{4+x^2}} \, \textrm{d} x \qquad \\[4mm] \textrm{d} u & = & 2 x \, \textrm{d} x \qquad & & v & = & \sqrt{4+x^2} \qquad \end{array} $$ we obtain $$ \int^1_0 \, \frac{x^2 \cdot x}{\sqrt{4+x^2}} \, \textrm{d} x \ {=} \big[ x^2 \cdot \sqrt{4 + x^2} \big]^1_0 \, - \, \int^1_0 \, 2x \cdot \sqrt{4 + x^2} \, \textrm{d} x \ {=} $$ $$ \ {=}\sqrt{5} \, - \, \Big[ \tfrac{2}{3} (4 + x^2)^{3/2} \Big]^1_0 \, \, = \, \, \tfrac{16}{3} - \tfrac{7}{3} \sqrt{5} $$

Exercise 6: Evaluating an Indefinite Integral

$$ {\displaystyle{ \int \, {\rm{e}}^x \sin(\pi x) \, \textrm{d} x }} $$

Hint (please, click on the "+" sign to read more) integration by parts twice may be necessary

Solution (please, click on the "+" sign to read more) $$ \ { } {\rm{e}}^x \sin(\pi x) - \pi {\rm{e}}^x \cos(\pi x) - \pi^2 \int \, {\rm{e}}^x \sin(\pi x) \, \textrm{d} x $$

Solution Path (please, click on the "+" sign to read more) by choosing $$ \begin{array}{r c l c r c l} u & = & \sin(\pi x) \qquad & \qquad \qquad & \textrm{d} v & = & {\rm{e}}^x \, \textrm{d} x \qquad \\[4mm] \textrm{d} u & = & \pi \, \cos(\pi x) \, \textrm{d} x \qquad & & v & = & {\rm{e}}^x \qquad \end{array} $$ we first obtain $$ \int \, {\rm{e}}^x \sin(\pi x) \, \textrm{d} x \ {=} {\rm{e}}^x \sin(\pi x) - \pi \int \, {\rm{e}}^x \cos(\pi x) \, \textrm{d} x $$ which makes second integration by parts necessary, so next by choosing $$ \begin{array}{r c l c r c l} u & = & \cos(\pi x) \qquad & \qquad \qquad & \textrm{d} v & = & {\rm{e}}^x \, \textrm{d} x \qquad \\[4mm] \textrm{d} u & = & -\pi \, \sin(\pi x) \, \textrm{d} x \qquad & & v & = & {\rm{e}}^x \qquad \end{array} $$ we obtain $$ \int \, {\rm{e}}^x \sin(\pi x) \, \textrm{d} x \ {=} \ { } {\rm{e}}^x \sin(\pi x) - \pi \int \, {\rm{e}}^x \cos(\pi x) \, \textrm{d} x \\[2mm] \ {=} $$ $$ \ {=} \ { }{\rm{e}}^x \sin(\pi x) - \pi \left( {\rm{e}}^x \cos(\pi x) + \pi \int \, {\rm{e}}^x \sin(\pi x) \, \textrm{d} x \right) \\[2mm] \ {=} $$ $$ \ {=} \ { } {\rm{e}}^x \sin(\pi x) - \pi {\rm{e}}^x \cos(\pi x) - \pi^2 \int \, {\rm{e}}^x \sin(\pi x) \, \textrm{d} x $$