Week 3 Activity

Section 3.4 Week 3 Activity

Subsection 3.4.1 Substitution Rule

This portion of the activity is direclty repeated from Calculus I. Do as much of it as you feel you need for review.

Activity 3.4.1.

Calculate this integral.

\begin{equation*} \int 3x^2 (x^3 - 4)^5 dx \end{equation*}

Solution.

The second part of this integrand looks like a composition: \((x^3 - 4)^5\) is an inside function, with an outside function \(u^5\text{.}\) Moreover, the remaining piece of the integral looks like it might be the derivative of the inside, which is idea for the substitution rule. Here is the substitution.

\begin{align*} u \amp = x^3 - 4 \\ du \amp = 3x^2 dx \end{align*}

I can rearrange the original integral to match these pieces, then do the replacement.

\begin{align*} \int 3x^2 (x^3 - 4)^5 dx \amp = \int (x^3-4)^5 (3x^2 dx) \\ \amp = \int u^5 du = \frac{u^6}{6} + c \end{align*}

At the end, I undo the substitution to return to the original variable.

\begin{equation*} \frac{u^6}{6} + c = \frac{(x^3-4)^6}{6} + c \end{equation*}

Activity 3.4.2.

Calculate this integral.

\begin{equation*} \int \cos (17 x) dx \end{equation*}

Solution.

The integrand is a composition with inside \((17x)\) and outside \(\cos (u)\text{.}\) We’ll try a substitution with these pieces.

\begin{align*} u \amp = 17 x \\ du \amp = 17 dx \\ \frac{1}{17} du \amp = dx \end{align*}

I had to move the 17 to the left, since the original integral doesn’t have a term which looks like \(17 dx\text{.}\) In this form, however, I can replace all the piece of the original integral with the new variable.

\begin{equation*} \int \cos (17 x) dx = \int \cos u \frac{1}{17} du \end{equation*}

Then I can use linearity and known antideriatives to do the integral.

\begin{equation*} = \frac{1}{17} \sin (u) + c \end{equation*}

Finally, I reverse the substitution.

\begin{equation*} = \frac{1}{17} \sin (17 x) + c \end{equation*}

Activity 3.4.3.

Calculate this integral.

\begin{equation*} \int \frac{x-4}{x-5} dx \end{equation*}

Solution.

This doesn’t look like a composition. However, substitution is sometime still useful even when I don’t obviously have a composition. Here, I’m going to take the denominator as the substitution, to try to produce an easier integral.

\begin{align*} u \amp = x - 5\\ du \amp = dx \end{align*}

I also need to know how to replace the numerator.

\begin{equation*} u = x - 5 \implies u + 1 = x - 4 \end{equation*}

Now I can replace all the piece of the original integral.

\begin{equation*} \int \frac{x-4}{x-5} dx = \int \frac{u+1}{u} du \end{equation*}

Now that I have a simpler denominator, I can split the numerator up to give two integrals, using linearity.

\begin{equation*} = \int \frac{u}{u} + \frac{1}{u} du = \int 1 du + \int \frac{1}{u} du \end{equation*}

Then I use known antiderivatives.

\begin{equation*} = u + \ln |u| + c \end{equation*}

Finally, I reverse the substitution.

\begin{equation*} = (x-5) + \ln |x-5| + c \end{equation*}

As a final note, I could group the \((-5)\) with the constant, like this:

\begin{equation*} = x + \ln |x-5| + (c - 5) \end{equation*}

If I did that, since the constant can be set to any real number, subtracting 5 doesn’t make a difference. Therefore, an alternative form of the answer gets ride of this \((-5)\text{.}\)

\begin{equation*} = x + \ln |x-5| + c \end{equation*}

This certinaly isn’t necessary, but I mention it here since these kinds of steps with constants of integration are commonly done.

Activity 3.4.4.

Calculate this integral.

\begin{equation*} \int_0^1 2x(x^2 + 4)^3 dx \end{equation*}

Solution.

This looks like a composition with inside function \((x^2+4)\) and outside function \(u^3\text{.}\) The rest of the integrand looks like it might be close to the derivative of the inside function, which is promising. We proceed with the substitution rule. Now I will also change the bounds.

\begin{align*} u \amp = x^2 + 4\\ du \amp = 2x dx\\ x \amp = 0 \implies u = (0)^2 + 4 = 4 \\ x \amp = 1 \implies u = (1)^2 + 4 = 5 \end{align*}

I can rearrange the integral so that I can substitute all the pieces clearly.

\begin{equation*} \int_0^1 2x(x^2 + 4)^3 dx = \int_0^1 (x^2+4)^3 (2x dx) \end{equation*}

Then I replace all the pieces, including changing the bounds.

\begin{equation*} = \int_4^5 u^3 du \end{equation*}

Then I can proceed to finish the integral. Since this is a definite integral and I have changed the bounds as well, I do not need to reverse the substitution.

\begin{equation*} = \frac{u^4}{4} \Big|_4^5 = \frac{5^4}{4} - \frac{4^4}{4} = \frac{625 - 256}{4} = \frac{369}{4} \end{equation*}

Activity 3.4.5.

Calculate this integral.

\begin{equation*} \int_0^4 \frac{x}{x+3} dx \end{equation*}

Solution.

I’d like to simplify the denominator, so I’ll try a substitution with \(u = (x+3)\text{.}\) Since this is a definite integral, I’ll change the bounds as well. Finally, I’ll need to replace the numerator in the substitution, so I’ll figure out how to express the numerator in the new variable.

\begin{align*} u \amp = x+3\\ du \amp = dx \\ x \amp = 0 \implies u = 0 + 3 = 3 \\ x \amp = 4 \implies u = 4 + 3 = 7 \\ u \amp = x + 3 \implies x = u - 3 \end{align*}

Now I have all the pieces and I can change to the new variable, including changing the bounds.

\begin{equation*} \int_0^4 \frac{x}{x+3} dx = \int_3^7 \frac{u-3}{u} du \end{equation*}

Then I can split up the integral using linearity into two integrals that I can solve with known antiderivatives.

\begin{align*} \amp = \int_3^7 \frac{u}{u} du - 3 \int_3^7 \frac{1}{u} du \\ \amp = \int_3^7 1 du - 3 \int_3^7 \frac{1}{u} du \\ \amp = u \Big|_3^7 - 3 \ln |u| \Big|_3^7 = (7-3) - 3 (\ln 7 - \ln 3) = 4 - 3 (\ln 7 - \ln 3) \end{align*}

Activity 3.4.6.

Calculate this integral.

\begin{equation*} \int_0^4 \frac{x}{x-3} \end{equation*}

Solution.

This look very similar to the previous question. One might thing that the process should be exactly the same. You could follow that process and get an answer, but that answer would not be valid. The problem, here, is that this is a definite integral, which calculates an area under the curve. Given these bounds, I am looking for the area between 0 and 4. However, this function is undefined at \(x=3\text{,}\) where it has a vertical asymptote. Therefore, the notion of area under this curve is meaningless. There is no connected curve under which I can define an area. This integral has no answer.

Subsection 3.4.2 Integration By Parts

Activity 3.4.7.

Solve this integral using integation by parts.

\begin{equation*} \int x \sin x dx \end{equation*}

Solution.

I choose \(\frac{df}{dx} = \sin x \) and \(g = x \text{.}\) The new functions are \(f = -\cos x \) and \(\frac{dg}{dx} = 1 \text{.}\) Then I apply integration by parts.

\begin{align*} \int x \sin x dx \amp = -x \cos x + \int \cos x dx \\ \amp = - x \cos x + \sin x + c \end{align*}

Activity 3.4.8.

Solve this integral using integation by parts.

\begin{equation*} \int x^3 \ln x dx \end{equation*}

Solution.

I choose \(\frac{df}{dx} = x^3\) and \(g = \ln x \text{.}\) The new functions are \(f = \frac{x^4}{4}\) and \(\frac{dg}{dx} = \frac{1}{x}\text{.}\) Then I apply integration by parts.

\begin{align*} \int x^3 \ln x dx \amp = \frac{x^4 \ln x}{4} - \int \frac{1}{x} \frac{x^4}{4} dx \\ \amp = \frac{x^4 \ln x}{4} - \frac{1}{4} \int x^3 dx = \frac{x^4 \ln x}{4} - \frac{x^4}{16} + c \end{align*}

Activity 3.4.9.

Solve this integral using integation by parts.

\begin{equation*} \int x^2 \sin x dx \end{equation*}

Solution.

I choose \(\frac{df}{dx} = \sin x\) and \(g = x^2 \text{.}\) The new functions are \(f = -\cos x\) and \(\frac{dg}{dx} = 2x\text{.}\) Then I apply integration by parts.

\begin{align*} \int x^2 \sin x dx \amp = -x^2 \cos x + \int 2x \cos x dx \end{align*}

Now I have to do integration by parts a second time on the remianing integral. I choose \(\frac{df}{dx} = \cos x\) and \(g = 2x \text{.}\) The new functions are \(f = \sin x\) and \(\frac{dg}{dx} = 2\text{.}\) Then I apply integration by parts.

\begin{align*} \int x^2 \sin x dx \amp = -x^2 \cos x + 2x \sin x - \int 2 \sin x dx \\ \amp = -x^2 \cos x + 2x \sin x + 2 \cos x + c \end{align*}

Activity 3.4.10.

Solve this integral using integation by parts.

\begin{equation*} \int_1^4 x \ln x dx \end{equation*}

Solution.

I choose \(\frac{df}{dx} = x\) and \(g = \ln x\text{.}\) The new functions are \(f = \frac{x^2}{2}\) and \(\frac{dg}{dx} = \frac{1}{x}\text{.}\) Then I apply integration by parts. Since this is a definite integral, I evalute the middle non-integral term on the bounds.

\begin{align*} \int_1^4 x \ln x dx \amp = \frac{ x^2 \ln x}{2} \Bigg|_1^4 - \int_1^4 \frac{1}{x} \frac{x^2}{2} dx \\ \amp = \frac{4^2 \ln 4}{2} - \frac{1^2 \ln 1}{2} - \frac{1}{2} \int_1^4 x dx \\ \amp = 8 \ln 4 - \frac{1}{2} \frac{x^2}{2} \Bigg|_1^4 = 8 \ln 4 - \frac{4^2}{4} + \frac{1}{4} = 8 \ln 4 - \frac{15}{4} \end{align*}

Activity 3.4.11.

Solve this integral using integation by parts. (This doesn’t look like integration by parts will really solve much. However, after you apply integration by parts, you will recover an integration that is the same as the original. Then use algebra to rearrange the equation and end up with a valid solution to the integral.)

\begin{equation*} \int \sin x \cos x dx \end{equation*}

Solution.

I choose \(\frac{df}{dx} = \cos x\) and \(g = \sin x\text{.}\) The new functions are \(f = \sin x \) and \(\frac{dg}{dx} = \cos x\text{.}\) Then I apply integration by parts.

\begin{equation*} \int \sin x \cos x dx = \sin^2 x - \int \sin x \cos x dx \end{equation*}

Now I can move the second term on the right to the left. Since is it exactly the same integral, when I add it to the left, I get two of them. Then I just divide by \(2\) for the final solution.

\begin{align*} 2 \int \sin x \cos x dx \amp = \sin^2 x \\ \int \sin x \cos x dx \amp = \frac{\sin^2 x}{2} + c \end{align*}

Activity 3.4.12.

Solve this integral using integation by parts.

\begin{equation*} \int x \sqrt{x+6} dx \end{equation*}

Solution.

I choose \(\frac{df}{dx} = \sqrt{x+6}\) and \(g = x\text{.}\) Seeing the derivative of \(g\) is relativel easy: \(\frac{dg}{dx} = 1\text{.}\) However, the antiderivative of \(\frac{df}{dx}\) requires more work. Let me actually do this integral.

\begin{equation*} \int \sqrt{x+6} dx \end{equation*}

I use the substitution \(u = x+6\) with \(du = dx\)

\begin{equation*} \int \sqrt{x+6} dx = \int \sqrt{u} du = \int u^{\frac{1}{2}} du = \frac{2}{3} u^{\frac{3}{2}} \end{equation*}

After undoing the substitution, I see the antiderivative is \(f = \frac{2}{3} (x+6)^{\frac{3}{2}}\text{.}\) Now I can apply integration by parts.

\begin{align*} \int x \sqrt{x+6} dx \amp = \frac{2x}{3} (x+6)^{\frac{3}{2}} - \int \frac{2}{3} (x+6)^{\frac{3}{2}} dx \\ \amp = \frac{2x}{3} (x+6)^{\frac{3}{2}} - \frac{2}{3} \frac{2}{5} (x+6)^{\frac{5}{2}} + c \\ \amp = \frac{2x}{3} (x+6)^{\frac{3}{2}} - \frac{4}{15} (x+6)^{\frac{5}{2}} + c \end{align*}

Activity 3.4.13.

Solve this integral using integation by parts. (This is a particularly long and challenging example which needs integration by parts twice and has a lot of difficult arithmetic.)

\begin{equation*} \int_0^1 x^2 \sqrt[3]{x+2} dx \end{equation*}

Solution.

I choose \(\frac{df}{dx} = \sqrt[3]{x+2}\) and \(g = x^2 \text{.}\) The antiderivative is \(f = \frac{3}{4} (x+2)^{\frac{4}{3}}\text{,}\) which is found by a substitution in a similar method to the previous question. The derivative is\(\frac{dg}{dx} = 2x\text{.}\) Then I apply integration by parts. Since this is a definite integral, I apply the bounds to the non-integral term.

\begin{align*} \int_0^1 x^2 \sqrt[3]{x+2} dx \amp = \frac{3x^2}{4} (x+2)^{\frac{4}{3}} \Bigg|_0^1 - \int_0^1 2x \frac{3}{4} (x+2)^{\frac{4}{3}} dx \\ \amp = \frac{3(1)^2}{4} 3^{\frac{4}{3}} - \frac{3(0)^2}{4} 2^{\frac{4}{3}} - \frac{3}{2}\int_0^1 x (x+2)^{\frac{4}{3}} dx \end{align*}

I need to do integration by parts again. I choose \(\frac{df}{dx} = (x+2)^{\frac{4}{3}}\) and \(g = x\text{.}\) The antiderivative is \(f = \frac{3}{7} (x+2)^{\frac{7}{3}}\text{,}\) again found by substitution and the inverse power rule. The derivative is\(\frac{dg}{dx} = 1\text{.}\) Then I apply integration by parts a second time. Again, since this is a definite integral, I apply the bounds to the non-integral term. The final integral that results is an inverse power rule integral. After evaluating the pieces on the bounds as I go along, I get a lot of pretty annoying arithmetic. I’ve tried to simplify it as best I can while still keeping it in exact values.

\begin{align*} \int_0^1 x^2 \sqrt[3]{x+2} dx \amp = \frac{3}{4} 3 \sqrt[3]{3} - \frac{3}{2} \left[ \frac{3x}{7} (x+2)^{\frac{7}{3}} \Bigg|_0^1 - \int_0^1 \frac{3}{7} (x+2)^{\frac{7}{3}} dx \right]\\ \amp = \frac{9\sqrt[3]{3}}{4} - \frac{3}{2} \left[ \frac{3(1)}{7} 3^{\frac{7}{3}} - \frac{3(0)}{7} 2^{\frac{7}{3}} - \frac{3}{7} \frac{3}{10} (x+2)^{\frac{10}{3}} \Bigg|_0^1 \right] \\ \amp = \frac{9\sqrt[3]{3}}{4} - \frac{3}{2} \left[ \frac{3}{7} 9 \sqrt[3]{3} - \frac{9}{70} 3^{\frac{10}{3}} + \frac{9}{70} 2^{\frac{10}{3}} \right]\\ \amp = \frac{9\sqrt[3]{3}}{4} - \frac{81\sqrt[3]{3}}{14} + \frac{27(27\sqrt[3]{3})}{140} - \frac{27(8\sqrt[3]{2})}{140} \\ \amp = \frac{(9(35) - 81(10) + 729)\sqrt[3]{3} - 216\sqrt[3]{2}}{140} = \frac{117\sqrt[3]{3} - 108\sqrt[3]{2}}{70} \end{align*}

Activity 3.4.14.

Solve this integral using integation by parts. (This is tricky, but you can use \(\frac{df}{dx} = 1\) in the integration by parts setup to actually make this work.)

\begin{equation*} \int (\ln x)^2 dx \end{equation*}

Solution.

I choose \(\frac{df}{dx} = 1\) and \(g = (\ln x)^2 \text{.}\) The new functions are \(f = \frac{2 \ln x}{x}\) and \(\frac{dg}{dx} = x\text{.}\) Then I apply integration by parts.

\begin{align*} \int (\ln x)^2 dx \amp = x (\ln x)^2 - \int \frac{2 \ln x}{x} x dx = x (\ln x)^2 - \int 2 \ln x dx \\ \amp = x (\ln x)^2 - 2 ( x \ln x - x) + c = x (\ln x)^2 - 2x \ln x + 2x + c \end{align*}

Activity 3.4.15.

Solve this integral using integation by parts.

\begin{equation*} \int x^2 \ln (x^2) dx \end{equation*}

Solution.

I choose \(\frac{df}{dx} = x^2\) and \(g = \ln (x^2) \text{.}\) The new functions are \(f = \frac{x^3}{3} \) and \(\frac{dg}{dx} = \frac{2}{x} \text{.}\) Then I apply integration by parts.

\begin{align*} \int x^2 \ln (x^2) dx \amp = \frac{x^3 \ln (x^2)}{3} - \int \frac{2}{x} \frac{x^3}{3} dx \\ \amp = \frac{x^3 \ln (x^2)}{3} - \frac{2}{3} \int x^2 dx \\ \amp = \frac{x^3 \ln (x^2)}{3} - \frac{2}{3} \frac{x^3}{3} + c = \frac{2x^3 \ln x}{3} - \frac{2x^3}{9} + c \end{align*}

Subsection 3.4.3 Conceptual Review Questions

What does it mean to change an integral from one variable to a new variable with substitution?
How is integration by parts a reverse of the product rule?

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