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In Exercises \(27-40\) , evaluate each integral using Part 2 of the Fundamental Theorem. Support your answer with NINT if you are unsure. $$\int_{0}^{\pi / 3} 2 \sec ^{2} \theta d \theta$$

Short Answer

Expert verified
The result is \(2 \sqrt{3}\).

Step by step solution

01

Evaluate Integral

The indefinite integral of \(2 \sec^2 \theta\) is \(2 \tan \theta\), since \(d/d \theta \tan \theta = \sec^2 \theta\).
02

Fundamental Theorem of Calculus Part 2

The Fundamental Theorem states that the integral of a function from a to b is its antiderivative evaluated at b minus its antiderivative evaluated at a, specifically \(\int_{a}^{b} f(x) dx = F(b) - F(a)\). Here, we substitute the limits of integration \(\theta = 0\) and \(\theta = \pi /3\) into the result from Step 1.
03

Calculate Result

Substitute \(\pi/3\) and \(0\) into \(2 \tan \theta\). We get \(2 \tan (\pi /3) - 2 \tan (0) = 2 \sqrt{3} - 0 = 2 \sqrt{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
A definite integral is a fundamental concept in calculus that represents the accumulation of quantities. It calculates the net area under a curve between two specified points. In mathematical terms, it is denoted as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration and \( f(x) \) is the integrand. The definite integral gives us the exact value of the net area, considering the integral of the function \( f(x) \) from point \( a \) to point \( b \).

Here's how it works: the integral sums up infinitely small slices of the area under the curve, which might be above or below the x-axis. Positive and negative areas are combined to provide the total net area. An important tool for easily computing definite integrals is the Fundamental Theorem of Calculus, which links the antiderivative of a function to its integral.
Trigonometric Integrals
Trigonometric integrals involve integrating functions that have trigonometric expressions as part of their integrands. These often include functions like \( \sin(x) \), \( \cos(x) \), \( \tan(x) \), and their reciprocal counterparts such as \( \sec(x) \) and \( \csc(x) \). One common strategy is to use known derivatives of trigonometric functions to reverse-engineer what the integral might be.

For example:
  • The integral of \( \sec^2(x) \) comes from the derivative of \( \tan(x) \), which is \( \sec^2(x) \).
  • Similarly, the integral of \( \cos(x) \) is \( \sin(x) \) (and vice versa) because the derivative of \( \sin(x) \) is \( \cos(x) \).
By understanding the relationships between these trigonometric functions and their derivatives, solving trigonometric integrals can become much easier. It's especially handy in cases like the given one: \( \int_{0}^{\pi/3} 2 \sec^2 \theta \, d\theta \), where knowing these relationships helps solve the integral efficiently.
Antiderivative
The antiderivative is a fancy term for the opposite of taking a derivative. It's also known as an "indefinite integral" but without specific bounds, absorbing the constant of integration. The antiderivative helps to "undo" the operation of differentiation, and it's often needed to solve definite integrals.

When we deal with a definite integral, such as the example \( \int_{0}^{\pi/3} 2 \sec^2 \theta \, d \theta \), finding the antiderivative is the first step. Using known derivatives, we can deduce the antiderivative of a function. For example, since the derivative of \( \tan(\theta) \) is \( \sec^2(\theta) \), the antiderivative of \( \sec^2(\theta) \) is \( \tan(\theta) \).

In our specific integral, \( 2 \tan(\theta) \) serves as the antiderivative of \( 2 \sec^2(\theta) \). You then substitute the upper and lower limits of the integral into the antiderivative to get the result. This is how the Fundamental Theorem of Calculus is applied, by evaluating the antiderivative at the boundaries and calculating the difference.

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Most popular questions from this chapter

In Exercises 1-6, (a) use the Trapezoidal Rule with n = 4 to approximate the value of the integral. (b) Use the concavity of the function to predict whether the approximation is an overestimate or an underestimate. Finally, (c) find the integral's exact value to check your answer. $$\int_{0}^{2} x^{3} d x$$

In Exercises 1-6, (a) use the Trapezoidal Rule with n = 4 to approximate the value of the integral. (b) Use the concavity of the function to predict whether the approximation is an overestimate or an underestimate. Finally, (c) find the integral's exact value to check your answer. $$\int_{0}^{4} \sqrt{x} d x$$

Average Daily Holding cost Solon Container receives 450 drums of plastic pellets every 30 days. The inventory function (drums on hand as a function of days) is \(I(x)=450-x^{2} / 2\) (a) Find the average daily inventory (that is, the average value of \(I(x)\) for the 30 -day period). (b) If the holding cost for one drum is \(\$ 0.02\) per day, find the average daily holding cost (that is, the per-drum holding cost times the average daily inventory).

Archimedes (287-212 B.C.), inventor, military engineer, physicist, and the greatest mathematician of classical times, discovered that the area under a parabolic arch like the one shown here is always two- thirds the base times the height. (a) Find the area under the parabolic arch \( y=6-x-x^2, -3 \leq x \leq 2 \) (b) Find the height of the arch. (c) Show that the area is two-thirds the base times the height.

Writing to Learn In Example 2 (before rounding) we found the average temperature to be 65.17 degrees when we used the integral approximation, yet the average of the 13 discrete temperatures is only 64.69 degrees. Considering the shape of the temperature curve, explain why you would expect the average of the 13 discrete temperatures to be less than the average value of the temperature function on the entire interval.

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