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Chapter 7: Problem 17

Sketch the region whose signed area is represented by the definite integral, and evaluate the integral using an appropriate formula from geometry, where needed. (a) \(\int_{0}^{3} x d x\) (b) \(\int_{-2}^{-1} x d x\) (c) \(\int_{-1}^{4} x d x\) (d) \(\int_{-5}^{5} x d x\)

Short Answer

Expert verified
(a) 4.5, (b) -1, (c) 7.5, (d) 0.

Step by step solution

01

Understand the Function and Limits

We are asked to evaluate definite integrals of the form \( \int_{a}^{b} x \, dx \). This requires finding the area under the line \( y = x \) between the limits \( a \) and \( b \). The line \( y = x \) is a straight line through the origin with a slope of 1.
02

Graph and Sketch Each Area

For each integral, sketch the line \( y = x \) and the region bounded by this line, the x-axis, and the vertical lines at \( x = a \) and \( x = b \). For instance, for part (a), sketch the area from the x-axis up to the line, between \( x = 0 \) and \( x = 3 \).
03

Calculate Area using Geometry for (a)

The area under \( y = x \) from \( x = 0 \) to \( x = 3 \) forms a right triangle with a base and height of 3. The area is \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 3 = 4.5 \).
04

Calculate Signed Area for (b)

The line segment from \( x = -2 \) to \( x = -1 \) lies below the x-axis, forming a triangle with a base 1 and height 2. This area is negative: \(-\frac{1}{2} \times 1 \times 2 = -1 \).
05

Calculate Signed Area for (c)

The region from \( x = -1 \) to \( x = 4 \) consists of two parts: a small triangle below the x-axis (base 1, height 1), and a larger triangle above the x-axis (base 4, height 4). Combined area is \((-\frac{1}{2} \times 1 \times 1) + (\frac{1}{2} \times 4 \times 4) = -0.5 + 8 = 7.5 \).
06

Calculate Signed Area for (d)

The line is symmetric about the origin, forming two equal triangles, one negative (left) and one positive (right). Since they cancel out, the entire area is \(0\).

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Key Concepts

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

Signed Area
In calculus, the term "signed area" refers to the concept of calculating the area under a curve considering the position relative to the x-axis. When a curve lies above the x-axis, the area is positive. Conversely, if the curve falls below the x-axis, the area becomes negative. This concept is integral to evaluating definite integrals where the curve crosses the x-axis.

Consider the definite integral \( \int_{-2}^{-1} x \ dx \). This involves the line \( y = x \) which is a straightforward example to visualize signed areas. The line lies below the x-axis between \( x = -2 \) and \( x = -1 \), forming a small triangular region. Since it's below the axis, the area of this triangle, calculated using geometry as \(-\frac{1}{2} \times 1 \times 2 = -1\), is considered negative.

Signed areas are essential to correctly interpreting whether an area is adding to or subtracting from the total integral value. Through understanding and exploring this concept, we see clearer connections between calculus and real-world scenarios involving accumulation and depletion.
Geometry in Calculus
Calculating areas using geometry in calculus is a vital skill. The use of simple geometric formulas simplifies the process of evaluating definite integrals, especially when dealing with straight lines like \( y = x \).

For instance, consider \( \int_{0}^{3} x \ dx \). The region under consideration forms a right triangle with the x-axis between \( x = 0 \) and \( x = 3 \), having both a base and height of 3. Utilizing the area formula for a triangle \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \), we find the area to be \( \frac{1}{2} \times 3 \times 3 = 4.5 \).

Implementing geometric principles in calculus not only aids in visualizing the problem but also helps simplify otherwise complex integrals, giving students a pragmatic approach to problem-solving. This approach connects abstract mathematical concepts to tangible, real-world surfaces and spaces, reducing intimidation and enhancing comprehension.
Line y = x
The line \( y = x \) is a central figure in understanding basic concepts of calculus including definite integrals. It's a simple, linear graph that passes through the origin with a slope of 1. This makes it a perfect demonstration model for studying fundamental topics like area under a curve and symmetry.

Analyzing \( \int_{-5}^{5} x \ dx \), the line's symmetry about the origin simplifies our understanding of cancellation and balance in calculus. The segments \( x = -5 \) to \( x = 0 \) and \( x = 0 \) to \( x = 5 \) form two triangles of equal area but opposite sign because one segment is below the x-axis while the other is above.

Hence, when these areas are summed, they cancel each other out, giving a total integral of 0. This highlights the beauty of symmetry in mathematics, showing how integral concepts can often be more about appreciating balance and symmetry rather than mere calculations.

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Most popular questions from this chapter

Let the velocity function for a particle that is at the origin nitially and moves along an \(s\) -axis be \(v(t)=t \ln (t+0.1)\). (a) Generate the velocity versus time curve, and use it to make a conjecture about the sign of the displacement over the time interval \(0 \leq t \leq 1\) (b) Use a CAS to find the displacement.

(a) Let \(f\) be an odd function; that is, \(f(-x)=-f(x) .\) Invent a theorem that makes a statement about the value of an integral of the form $$\int_{-a}^{a} f(x) d x$$ (b) Confirm that your theorem works for the integrals \(\int_{-1}^{1} x^{3} d x \quad\) and \(\quad \int_{-\pi / 2}^{\pi / 2} \sin x d x\) (c) Let \(f\) be an even function; that is, \(f(-x)=f(x) .\) Invent a theorem that makes a statement about the relationship between the integrals $$\int_{-a}^{a} f(x) d x \quad \text { and } \quad \int_{0}^{a} f(x) d x$$ (d) Confirm that your theorem works for the integrals$$\int_{-1}^{1} x^{2} d x \quad \text { and } \quad \int_{-\pi / 2}^{\pi / 2} \cos x d x$$

express \(F(x)\) in a piecewise form that does not involve an integral. $$\frac{d y}{d x}=\sqrt[3]{x} ; y(1)=2$$

Define \(F(x)\) by $$F(x)=\int_{1}^{x}\left(t^{3}+1\right) d t$$ (a) Use Part 2 of the Fundamental Theorem of Calculus to find \(F^{\prime}(x)\) (b) Check the result in part (a) by first integrating and then differentiating.

Evaluate the definite integral by expressing it in terms of \(u\) and evaluating the resulting integral using a formula from geometry. $$\int_{\pi / 3}^{\pi / 2} \sin \theta \sqrt{1-4 \cos ^{2} \theta} d \theta ; u=2 \cos \theta$$
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