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Section 1.1 Section Title

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Figure 1.1.1. Solution to a differential equation
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Figure 1.1.2. Limit example

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Checkpoint 1.1.3.

Evaluate the definite integral \(\int_{0}^{\pi/3} \sin^{27} \theta \cos \theta \, d\theta\text{.}\)

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Solution.

Let \(u = \sin \theta\text{,}\) then \(du = \cos \theta \, d\theta\text{.}\) The limits of integration change as follows:

When \(\theta = 0\text{,}\) \(u = \sin(0) = 0\text{.}\)

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When \(\theta = \pi/3\text{,}\) \(u = \sin(\pi/3) = \sqrt{3}/2\text{.}\)

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\begin{align*} \int_{0}^{\sqrt{3}/2} u^{27} \, du \amp = \left. \frac{u^{28}}{28} \right|_{0}^{\sqrt{3}/2}\\ \amp = \frac{(\sqrt{3}/2)^{28}}{28} - \frac{0^{28}}{28}\\ \amp = \frac{3^{14}}{28 \cdot 2^{28}} \end{align*}

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Checkpoint 1.1.4.

For the function \(f(x,y) = x^2 + xy + y^2\text{,}\) find the gradient vector \(\nabla f(1,3)\text{.}\)

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Solution.

First, we find the partial derivatives:

\begin{align*} f_x \amp = 2x + y\\ f_y \amp = x + 2y \end{align*}

So, \(\nabla f(x,y) = \langle 2x + y, x + 2y \rangle\text{.}\)

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Evaluating at the point \((1,3)\text{:}\)

\begin{equation*} \nabla f(1,3) = \langle 2(1) + 3, 1 + 2(3) \rangle = \langle 5, 7 \rangle \end{equation*}

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Checkpoint 1.1.5.

Evaluate the double integral \(\int_{0}^{1} \int_{0}^{2-2y} (3xy - x^2 - y^2 + 6) \, dx \, dy\text{.}\)

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Solution.

Integrating with respect to \(x\) first:

\begin{align*} \int_{0}^{1} \int_{0}^{2-2y} (3xy - x^2 - y^2 + 6) \, dx \, dy \amp = \int_{0}^{1} \left[ \frac{3}{2}x^2y - \frac{x^3}{3} - xy^2 + 6x \right]_{0}^{2-2y} \, dy\\ \amp = \int_{0}^{1} \left( \frac{32}{3}y^3 - 22y^2 + 2y + \frac{28}{3} \right) \, dy\\ \amp = \left[ \frac{8}{3}y^4 - \frac{22}{3}y^3 + y^2 + \frac{28}{3}y \right]_{0}^{1}\\ \amp = \frac{8}{3} - \frac{22}{3} + 1 + \frac{28}{3} = \frac{17}{3} \end{align*}

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