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Chapter 8: Problem 3

Show that the area under the graph of the linear function \(f(x)=x\) between \(x=a\) and \(x=b(a

Short Answer

Expert verified
The area is \(\frac{1}{2}(b^2 - a^2)\).

Step by step solution

01

Understanding the Problem

We need to find the area under the graph of the function \(f(x) = x\) within the interval \([a, b]\). The problem asks us to show this area as \(\frac{1}{2}(b^2 - a^2)\).
02

Setup the Problem with Integration

The area under the curve of a function \(f(x)\) from \(x = a\) to \(x = b\) is given by the definite integral \(\int_{a}^{b} f(x) \, dx\). For our linear function, this becomes \(\int_{a}^{b} x \, dx\).
03

Integrate the Function

To solve \(\int_{a}^{b} x \, dx\), we first need to find the antiderivative of \(x\). The antiderivative of \(x\) is \(\frac{x^2}{2}\).
04

Apply the Fundamental Theorem of Calculus

Using the Fundamental Theorem of Calculus, we evaluate the antiderivative \(\frac{x^2}{2}\) at \(b\) and \(a\), and then find the difference: \[\int_{a}^{b} x \, dx = \left[ \frac{x^2}{2} \right]_a^b = \frac{b^2}{2} - \frac{a^2}{2}.\]
05

Simplify the Expression

The expression \(\frac{b^2}{2} - \frac{a^2}{2}\) simplifies to \(\frac{1}{2}(b^2 - a^2)\). Thus, the area under the function \(f(x) = x\) from \(x = a\) to \(x = b\) is \(\frac{1}{2}(b^2 - a^2)\).

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Key Concepts

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

Area Under a Curve
To find the area under a curve, we often rely on integration. Specifically, the area between a curve defined by a function and the x-axis over a certain interval can be calculated using a definite integral.
For the function \( f(x) = x \) on the interval \([a, b]\), we want to compute the exact area captured between the line and the x-axis.
  • This area calculation gives us a geometrical representation.
  • We locate it between the vertical lines intersecting the curve at \( x = a \) and \( x = b \).
  • Visually, it forms a trapezoid-like shape called a planar region.
The mathematical formula for this area is expressed as \( \int_{a}^{b} x \, dx \).Using integration, we derive a precise numeric value representing the size of this area, providing essential insights into the behavior of the curve across the interval we are interested in.
Linear Functions
Linear functions are the simplest type of polynomial functions, shaped like straight lines. The general form of a linear function is \( f(x) = mx + c \), where \( m \) is the slope of the line, and \( c \) is the y-intercept. In our particular problem, the linear function is \( f(x) = x \).
  • The slope \( m = 1 \), meaning the line rises one unit vertically for each unit it moves horizontally.
  • The y-intercept \( c = 0 \), indicating the line passes through the origin \((0, 0)\).
Understanding these characteristics allows us to deduce the fundamental properties of the function:
  • The rate of change is constant 鈥� hence, a constant slope.
  • The graph of \( f(x) = x \) is a perfect 45-degree diagonal across the plane.
This simplicity provides a unique advantage when computing areas under linear functions, as the resulting integration tends to be straightforward.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus beautifully links the concepts of differentiation and integration. It states that if we have a continuous function on an interval, we can find the total accumulation of that function over the interval using its antiderivative.For our case, the function is \( f(x) = x \), and the antiderivative, or primitive function, is \( \frac{x^2}{2} \).
  • The theorem tells us to evaluate this antiderivative at the endpoints of our interval.
  • So, we calculate \( \left[ \frac{x^2}{2} \right]_a^b \), which translates to \( \frac{b^2}{2} - \frac{a^2}{2} \).
By finding this difference, we determine the net area under the curve \( f(x) = x \) from \( x = a \) to \( x = b \).The simplicity of this result showcases the power of the Fundamental Theorem of Calculus in providing a direct way to compute definite integrals efficiently.

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Most popular questions from this chapter

Find the slope of the tangent to the lemniscate $$ \left(x^{2}+y^{2}\right)^{2}-a^{2}\left(x^{2}-y^{2}\right) $$ at the point \((x, y)\). (See Review exercises \(2.12\), Question 19.)

The displacement-time graph for a vehicle is given by $$ s(t)= \begin{cases}t, & 0 \leq t \leq 1 \\ t^{2}-t+1, & 1 \leq t \leq 2 \\ 3 t-3, & 2 \leq t=3 \\ 9-t, & 3 \leqslant t=9\end{cases} $$ Obtain the formula for the velocity-time graph.

Using partial fractions, integrate (a) \(\frac{x}{x^{2}-3 x-4}\) (b) \(\frac{x}{(x-2)^{2}}\) (c) \(\frac{1}{x(x+1)}\) (d) \(\frac{x}{x^{2}+2 x+1}\) (e) \(\frac{1}{x^{2}-1}\) (f) \(\frac{1}{x^{2}(x-1)}\) (g) \(\frac{1}{x(x-1)(x-2)}\) (h) \(\frac{1}{1+x-2 x^{2}}\) (i) \(\frac{2 x^{3}}{x^{3}-1}\) (j) \(\frac{3 x^{3}-3 x^{2}+4 x-2}{x(x-1)\left(x^{2}+1\right)}\) (k) \(\frac{9}{(x-1)(x+2)^{2}}\) (1) \(\frac{x^{2}-2 x+3}{(x-1)\left(x^{2}-x-1\right)}\)

Differentiate with respect to \(x\) : (a) \(\sinh ^{-1} 2 \pi\) (b) \(\cosh ^{-1}\left(2 x^{2}-1\right)\) (c) \(\tanh ^{-1}(1 / x)\) (d) \(\sqrt{\left(1+x^{2}\right) \sinh ^{-1} x}\) (e) \(\sqrt{\left(4-x^{2}\right)-2 \cosh ^{-1}(2 / x)}\) (f) \(\tanh ^{-1} x /\left(1+x^{2}\right)\)

Show that the radius of curvature at the origin to the curve $$ x^{3}+y^{3}+2 x^{2}-4 y+3 x=0 $$ is \(\frac{125}{64}\).
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