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Chapter 5: Problem 25

Evaluate the integral. $$\int_{0}^{\pi / 4} 2 \sec ^{2} x d x$$

Short Answer

Expert verified
The value of the integral \(\int_{0}^{\pi / 4} 2 \sec ^{2} x d x\) is 2.

Step by step solution

01

Identify the integral

The integral that needs to be solved is \(\int_{0}^{\pi / 4} 2 \sec ^{2} x d x\).
02

Evaluate the indefinite integral

We first want to compute the indefinite integral of \(2\sec^2x\) which is a standard result and can be found in common integral tables. The indefinite integral of \(2\sec^2x\) is \(2\tan x + C\), where \(C\) represents the constant of integration.
03

Evaluate the definite integral

To evaluate the definite integral, we compute \(F(b) - F(a)\), where \(F(x)\) represents the antiderivative of the integrated function, and \(a\) and \(b\) represent the lower and upper limits of integration. Substituting upper and lower limits gives \((2\tan(\pi/4)) - (2\tan(0))\).
04

Calculation

The values of \(\tan(\pi/4)\) and \(\tan(0)\) are known to be \(1\) and \(0\) respectively. Substituting these values in gives \(2*1 - 2*0 = 2\).

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Key Concepts

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

Definite Integrals
Definite integrals help us calculate the exact area under a curve between two endpoints. In this exercise, the definite integral \( \int_{0}^{\pi / 4} 2 \sec^2 x \, dx \) represents the area under the function \( 2 \sec^2 x \) from \( x = 0 \) to \( x = \pi / 4 \). Here鈥檚 a straightforward breakdown of how we handle definite integrals:
  • Identify the function to integrate and the limits of integration鈥攖he starting and ending points.
  • Find the antiderivative (or the indefinite integral) of the function. This provides a formula representing the accumulated area function.
  • Apply the Fundamental Theorem of Calculus, which tells us to evaluate this antiderivative at the upper limit and subtract the evaluation at the lower limit: \( F(b) - F(a) \). In this context, it鈥檚 \( F(\pi / 4) - F(0) \).
  • Compute these values to find the definite integral's result, offering a precise area measure between the two limits.
By following these steps, solving definite integrals becomes a straightforward process, even when dealing with trigonometric or more complex functions.
Trigonometric Functions
Trigonometric functions are crucial in integral calculus, often appearing in problems involving wave-like patterns or oscillations. In this exercise, we are tackling \( \sec^2 x \), which is the trigonometric function
  • \( \sec x \) is the reciprocal trigonometric function of \( \cos x \), so \( \sec x = \frac{1}{\cos x} \).
  • The function \( \sec^2 x \) is essential because it is the derivative of \( \tan x \). This fact is key in efficiently finding integrals and antiderivatives involving \( \sec^2 x \).
Understanding the interplay of these functions allows us to quickly identify integrals, as we did here: recognizing \( \int 2 \sec^2 x \, dx = 2 \tan x + C \) thanks to
  • The integral relationship: since the derivative of \( \tan x \) leads directly to \( \sec^2 x \), integrating \( \sec^2 x \) brings us back to \( \tan x \).
  • Function familiarity: knowing the basics of trigonometry helps in comprehending transformations and substitutions, crucial for solving integrals.
Antiderivatives
Antiderivatives, sometimes called indefinite integrals, reverse the process of differentiation. Essentially, if you have a derivative, finding its antiderivative lets you reconstruct the original function. In this problem, our task involved:
  • Identifying the function to integrate, which was \( 2 \sec^2 x \).
  • Finding the antiderivative, \( 2 \tan x + C \). We knew from calculus tables and understanding trigonometric derivatives that the derivative of \( \tan x \) is \( \sec^2 x \).
  • Applying the constant of integration \( C \), which is crucial for indefinite integrals to represent the family of functions that differ by a constant.
When solving definite integrals, this constant \( C \) isn't needed because it cancels out in evaluation. Remember,
  • Antiderivatives let you glimpse into how changes accumulate, similar to tracking savings after a steady contribution over time.
  • They're foundational in physics and engineering, providing insights into movement, area, and total accumulated quantities over intervals.

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Most popular questions from this chapter

Calculate. $$\frac{d}{d x}\left(\int_{1}^{\sqrt{x}} \frac{t^{2}}{1+t^{2}} d t\right)$$

Let \(P=\left(x_{0}, x_{1}, x_{2}, \ldots, x_{x-1}, x_{n}\right]\) be a regular partition of the interval \([0, b],\) and set \(f(x)=x\) (a) Show that $$L_{f}(P)=\frac{b^{2}}{n^{2}}(0+1+2+3+\cdots+(n-1)].$$ (b) Show that $$U_{f}(P)=\frac{b^{2}}{n^{2}}[1+2+3+\cdots+n].$$ (c) Use Exercise 35 to show that $$L_{f}(P)=\frac{1}{2} b^{2}(1-\|P\|) \text { and } U_{f}(P)=\frac{1}{2} b^{2}(1+\|P\|).$$ (d) Show that for all choices of \(x\);-poinis $$\lim _{\| P_{1} \rightarrow 0} S^{*}(P)=\frac{1}{2} b^{2} \quad \text { and therefore } \quad \int_{0}^{b} x d x=\frac{1}{2} b^{2}.$$

Using a regular partition \(P\) with 10 subintervals, estimate the integral (a) \(\operatorname{by} L_{f}(P)\) and by \(U_{f}(P),\) (b) by \(\frac{1}{2}\left[L_{f}(P)+U_{f}(P)\right]\) (c) by \(S^{-}(P)\) using the midpoints of the subintervals. How docs this result compare with your result in part (b)? $$\int_{0}^{2} \frac{1}{1+x^{2}} d x$$.

Calculate. $$\frac{d}{d x}\left(\int_{x}^{a} f(t) d t\right)$$

A particle moves along the \(x\) -axis with velocity \(v(t)=\) \(A t^{2}+1 .\) Determine \(A\) given that \(x(1)=x(0) .\) Compute the total distance traveled by the particle during the first second.
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