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PAGE 1

Definite Integrals and Area

Find the area of the region between the graph of \( y = \frac{1}{1+x^2} \) and the \( x \)-axis, from \( x = -\sqrt{3} \) to \( x = 1 \).

A. \( \frac{\pi}{2} \)B. \( \frac{3\pi}{4} \)C. \( \frac{15\pi}{12} \)D. \( \frac{\pi}{3} \)E. \( \frac{7\pi}{12} \)

\( \int_{a}^{b} f(x) dx \rightarrow \) area of region bounded by \( f(x) \) and \( x \)-axis from \( x=a \) to \( x=b \)

Solution Steps

Here, \( \int_{-\sqrt{3}}^{1} \frac{1}{1+x^2} dx \)

\( f(x) \rightarrow F(x) = ? \)

\( = \tan^{-1}(x) \Big|_{-\sqrt{3}}^{1} \)

\( = \tan^{-1}(1) - \tan^{-1}(-\sqrt{3}) \)

\( = \frac{\pi}{4} - (-\frac{\pi}{3}) \)

\( = \frac{\pi}{4} + \frac{\pi}{3} = \frac{7\pi}{12} \)

FTC 2:

\( \int_{a}^{b} f(x) dx = F(b) - F(a) \)

where \( F'(x) = f(x) \)

\( F(x) \) is antiderivative of \( f(x) \)

\( \theta = \tan^{-1}(-\sqrt{3}) \)

\( \tan \theta = -\sqrt{3} \)

\( = \frac{-\sqrt{3}/2}{1/2} \)

\( \theta = -\frac{\pi}{3} \)

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\( \int_{0}^{1} \frac{e^x}{1+e^x} dx = \)

A. \( \ln \frac{1+e}{2} \)B. \( \ln(1+e) \)C. \( \frac{1}{2} \)D. \( 1 - \ln 2 \)E. \( e \)

Integration by Substitution

\( \int_{0}^{1} \frac{e^x}{1+e^x} dx \)

\( e^x \) vs. \( 1+e^x \)

Find one part whose derivative is a constant multiple of the other part.

Rule of thumb: the more complicated part.

Here, \( \frac{d}{dx}(1+e^x) = e^x \rightarrow \) exactly the other part.

So, let \( u = 1+e^x \)

\( \frac{du}{dx} = e^x \implies du = e^x dx \)

Changing Limits of Integration

  • Old upper limit: \( x = 1 \rightarrow u = 1+e^1 \rightarrow u = 1+e \)
  • Old lower limit: \( x = 0 \rightarrow u = 1+e^0 \rightarrow u = 1+1 = 2 \)

\( \int_{2}^{1+e} \frac{1}{u} du = \ln |u| \Big|_{2}^{1+e} \)

\( = \ln |1+e| - \ln |2| \)

\( = \ln \frac{1+e}{2} \)

do NOT go back to x

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Integration by Substitution

Compute \[ \int_{\ln \frac{\pi}{4}}^{\ln \frac{\pi}{2}} e^x \cos e^x dx \]

  1. A. 1
  2. B. \(\frac{\sqrt{3} + 1}{2}\)
  3. C. \(1 - \sqrt{2}\)
  4. D. \(\frac{\sqrt{3} - 1}{2}\)
  5. E. \(\frac{2 - \sqrt{2}}{2}\)
\[ \int_{\ln \frac{\pi}{4}}^{\ln \frac{\pi}{2}} e^x \cos e^x dx \]

Analysis

Look at parts: \(e^x\), \(\cos e^x\)

Neither has a derivative that is a constant multiple of the other.

But if we focus on the two \(e^x\), notice if \(u = e^x\) then we can substitute cleanly.

\(u = e^x\)

\(\frac{du}{dx} = e^x\)

\(du = e^x dx\)

\[ \underbrace{\cos(e^x)}_{\cos(u)} \cdot \underbrace{e^x dx}_{du} \]

Adjusting Limits

\(x = \ln \frac{\pi}{2} \rightarrow u = e^x = e^{\ln \frac{\pi}{2}} = \frac{\pi}{2}\)

\(x = \ln \frac{\pi}{4} \rightarrow u = e^x = e^{\ln \frac{\pi}{4}} = \frac{\pi}{4}\)

New Integral

\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos(u) du = \sin(u) \Big|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{4}\right) = 1 - \frac{1}{\sqrt{2}} = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2} \]
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Fundamental Theorem of Calculus (FTC 2)

Compute \[ \int_{0}^{\frac{\pi}{2}} \frac{d}{dx}(x^4 \sin(x)) dx \]

  1. A. \(\frac{\pi}{2}\)
  2. B. 0
  3. C. 1
  4. D. \(\frac{\pi^3}{8}\)
  5. E. \(\frac{\pi^4}{16}\)

FTC 2:

\[ \int_{a}^{b} f(x) dx = F(b) - F(a) \]

where \(F' = f(x)\) or \(F\) is antiderivative of \(f(x)\)

Solution

\(f(x) = \frac{d}{dx}(x^4 \sin(x))\)

\(F(x)\) is antiderivative of \(\frac{d}{dx}(x^4 \sin(x))\)

so \(F(x) = x^4 \sin(x)\)

\[ \int_{0}^{\frac{\pi}{2}} \frac{d}{dx}(x^4 \sin(x)) dx = x^4 \sin(x) \Big|_{0}^{\frac{\pi}{2}} \]
\[ = \left(\frac{\pi}{2}\right)^4 \sin\left(\frac{\pi}{2}\right) - 0 \]
\[ = \frac{\pi^4}{16} \]
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Fundamental Theorem of Calculus: Derivative on the Outside

What if the derivative is on the outside?

\[ \frac{d}{dx} \left[ \int_{0}^{\pi/2} x^4 \sin(x) \, dx \right] \]

The integral expression inside the brackets represents the area between \( x^4 \sin(x) \) and the x-axis from \( x = 0 \) to \( x = \pi/2 \), which is always a number (constant).

\[ \frac{d}{dx} (\text{constant}) = 0 \]

Fundamental Theorem of Calculus Part 1 (FTC 1)

\[ \text{FTC 1: } \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \]

Note: One of the limits of integration has to be a non-constant (like \( x \)) for the derivative to be non-zero.

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S2018 #4: Indeterminate Forms and L'Hospital's Rule

Problem Statement

\[ \lim_{x \to \infty} \left( 1 + \frac{1}{2x} \right)^{3x} \xrightarrow{x \to \infty} 1^{\infty} \text{ (indeterminate form)} \]

We need to turn it into \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) then use L'Hospital's Rule.

Step 1: Logarithmic Transformation

Let \( y = \left( 1 + \frac{1}{2x} \right)^{3x} \). We want to find \( \lim_{x \to \infty} y \).

\[ \ln y = \ln \left( 1 + \frac{1}{2x} \right)^{3x} \]\[ \ln y = 3x \cdot \ln \left( 1 + \frac{1}{2x} \right) = \frac{\ln \left( 1 + \frac{1}{2x} \right)}{\frac{1}{3x}} \]

Step 2: Check for Indeterminate Form

\[ \frac{\ln \left( 1 + \frac{1}{2x} \right)}{\frac{1}{3x}} \xrightarrow{x \to \infty} \frac{\ln(1) = 0}{0} \to \frac{0}{0} \]

Now we can evaluate the limit of the natural log:

\[ \lim_{x \to \infty} \ln y = \lim_{x \to \infty} \frac{\ln \left( 1 + \frac{1}{2x} \right)}{\frac{1}{3x}} \]
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Limit Evaluation and Logarithmic Substitution

\[ \lim_{x \to \infty} \frac{\frac{1}{1+\frac{1}{2x}} \frac{d}{dx}(1+\frac{1}{2x})}{\frac{d}{dx}(\frac{1}{3x})} = \lim_{x \to \infty} \frac{\frac{1}{1+\frac{1}{2x}} \cdot \frac{1}{2} \cdot \frac{-1}{x^2}}{\frac{1}{3} \cdot \frac{-1}{x^2}} \]

Note on the side:

\[ \frac{d}{dx}(\frac{1}{2x}) = \frac{1}{2} \frac{d}{dx}(\frac{1}{x}) = \frac{1}{2} \cdot \frac{-1}{x^2} \]

Simplifying the limit expression by canceling common terms:

\[ = \lim_{x \to \infty} \frac{\frac{1}{2} \cdot \frac{1}{1+\frac{1}{2x}}}{\frac{1}{3}} = \frac{\frac{1}{2}}{\frac{1}{3}} = \frac{3}{2} \]

Note: As \( x \to \infty \), the term \( \frac{1}{2x} \to 0 \).

Final Result for y

We found that:

\[ \lim_{x \to \infty} \ln y = \frac{3}{2} \quad \text{but we want } \lim_{x \to \infty} y \]

Using the property \( y = e^{\ln y} \):

\[ \text{so } \lim_{x \to \infty} y = \lim_{x \to \infty} e^{\ln y} = e^{3/2} \]
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S19 # 8

Finding the Derivative of a Logarithmic Function

Find \( f'(1) \) if \( f(x) = \ln\left(\frac{x}{x^2+2}\right) \)

One way to do this:

\[ f'(x) = \frac{1}{\frac{x}{x^2+2}} \cdot \frac{d}{dx}\left(\frac{x}{x^2+2}\right) \]

This requires the quotient rule. It's ok but might get messy.

Another way:

Use logarithmic properties first:

\[ f(x) = \ln\left(\frac{x}{x^2+2}\right) = \ln(x) - \ln(x^2+2) \]

Now each term has a simple derivative.

\[ f'(x) = \frac{1}{x} - \frac{1}{x^2+2} \cdot \frac{d}{dx}(x^2+2) \]\[ = \frac{1}{x} - \frac{2x}{x^2+2} \]

Evaluating at \( x = 1 \):

\[ f'(1) = \frac{1}{1} - \frac{2}{1+2} = 1 - \frac{2}{3} = \frac{1}{3} \]
PAGE 9

Derivative of an Area Function

\[ f(x) = \int_{3}^{\tan x} \sqrt{\sqrt{t} + 6t} \, dt \]

Find \( f'(\frac{\pi}{4}) \).

Derivative of Area Function: FTC 1

\[ \text{FTC 1} : \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \]

If \( x \) is not just \( x \), use Chain Rule.

\[ f(x) = \int_{3}^{\tan x} \sqrt{\sqrt{t} + 6t} \, dt \]

Let \( u = \tan x \), then

\[ f(u) = \int_{3}^{u} \sqrt{\sqrt{t} + 6t} \, dt \]

Chain Rule:

\[ \frac{df}{dx} = \frac{df}{du} \frac{du}{dx} \]
\[ \frac{df}{dx} = \left( \underbrace{\frac{d}{du} \int_{3}^{u} \sqrt{\sqrt{t} + 6t} \, dt}_{\text{FTC 1}} \right) \left( \frac{d}{dx} \tan x \right) \]
\[ = \left( \sqrt{\sqrt{u} + 6u} \right) (\sec^2 x) = \left( \sqrt{\sqrt{\tan x} + 6 \tan x} \right) (\sec^2 x) \]
PAGE 10

Evaluation at \( x = \frac{\pi}{4} \)

At \( x = \frac{\pi}{4} \):

  • \( \tan x = 1 \)
  • \( \sec x = \frac{1}{\cos x} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \)

So,

\[ f'\left(\frac{\pi}{4}\right) = \sqrt{\sqrt{1} + 6} \cdot (\sqrt{2})^2 = 2\sqrt{7} \]