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Chapter 5: Problem 26

Find the area under the curve \(y=1 /(3 x+1)^{2}\) over the interval [0,1].

Short Answer

Expert verified
The area under the curve is \(\frac{1}{4}\).

Step by step solution

01

Set up the Integral

To find the area under the curve \(y = \frac{1}{(3x+1)^2}\) over the interval \([0,1]\), we need to set up the definite integral: \[ \int_{0}^{1} \frac{1}{(3x+1)^2} \; dx. \]
02

Use Substitution Method

Let \(u = 3x + 1\), which implies \(du = 3\,dx\), or \(dx = \frac{du}{3}\). Change the limits of integration: \(x = 0\) gives \(u = 1\), and \(x = 1\) gives \(u = 4\). Substitute to reformulate the integral: \[ \int_{1}^{4} \frac{1}{u^2} \cdot \frac{1}{3} \; du. \]
03

Simplify the Integral

Factor out the constant \(\frac{1}{3}\) from the integral: \[ \frac{1}{3} \int_{1}^{4} \frac{1}{u^2} \; du. \] Recall that \(\frac{1}{u^2}\) can be rewritten as \(u^{-2}\), so the integral becomes: \[ \frac{1}{3} \int_{1}^{4} u^{-2} \; du. \]
04

Integrate

Use the power rule for integration: \(\int u^n \, du = \frac{u^{n+1}}{n+1} + C\), where \(n eq -1\). Here \(n = -2\), so: \[ \int u^{-2} \, du = \frac{u^{-1}}{-1} = -\frac{1}{u}. \] Apply the evaluated expression: \[ \frac{1}{3} \left[ -\frac{1}{u} \right] \] evaluated from 1 to 4.
05

Evaluate the Definite Integral

Substitute the limits of integration into the expression: \[ \frac{1}{3} \left( -\frac{1}{4} - \left(-\frac{1}{1}\right) \right) = \frac{1}{3} \left( -\frac{1}{4} + 1 \right). \] Simplify: \[ \frac{1}{3} \left( \frac{3}{4} \right) = \frac{1}{4}. \] Thus, the area under the curve is \(\frac{1}{4}\).

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Key Concepts

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

Definite Integral
The concept of a definite integral is central in calculus, as it provides a way to calculate the area under a curve over a specific interval. Basically, it takes a function and computes the total accumulation of that function across a range of inputs. For example, if you want to find the area under a curve from point A to point B on the x-axis, a definite integral comes into play. The integral is set up with limits, in this case from 0 to 1, to cover just the portion of the curve where you need the area computed.
  • A definite integral is often expressed with limits, like \ \int_a^b f(x) \, dx \, where \(a\) and \(b\) are the limits of integration.
  • It provides a scalar value that represents the area under the curve between these limits.
  • The integral is evaluated to produce a precise numeric answer, revealing crucial area information.
Understanding the setup and application of a definite integral is key to solving such calculus problems effectively.
Substitution Method
The substitution method simplifies complex integrals by making a change of variable, which reshapes the integral into a more manageable form. This technique is particularly useful when dealing with complicated functions, such as those involving composite functions.In the substitution method, you often let a new variable \(u\) equal a function of \(x\). For the integral \(\int \frac{1}{(3x+1)^2} \, dx\), the substitution \(u = 3x + 1\) makes the integral easier to handle. Doing this also requires adjusting the differential \(dx\) and recalibrating the integration limits according to \(u\).
  • Substitute: Choose a substitution that simplifies the integral; here, \(u = 3x + 1\) means \(dx = \frac{du}{3}\).
  • Recalculate limits: Transform the original limits from \(x\) to \(u\); when \(x = 0\), \(u = 1\) and when \(x = 1\), \(u = 4\).
  • Integrate: Solve the new integral in terms of \(u\), then switch back to the original variable, if necessary.
Mastering substitution allows you to tackle tricky integrals by aligning them with easier forms.
Area Under a Curve
Calculating the area under a curve is a common calculus problem that makes use of definite integrals. This technique is valuable for finding the total amount of a quantity that can vary, such as distance traveled or total income over a period of time.To find the area under a curve like \(y = \frac{1}{(3x + 1)^2}\), you create and evaluate a definite integral over the defined interval, such as from x = 0 to x = 1. The process uses calculus tools to estimate what might be irregular or curved shapes:
  • Start by expressing the curve with a function \(f(x)\) that defines how y changes with x.
  • Integrate this function over the chosen interval, replacing a simple geometric area calculation with the more nuanced integral.
  • The definite integral result quantifies the precise area under the line of \(y = f(x)\), providing a direct answer.
Finding areas under curves is practical, giving insights into accumulative measures in real-world scenarios.

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Most popular questions from this chapter

(a) Evaluate the integral \(\int(5 x-1)^{2} d x\) by two methods: first square and integrate, then let \(u=5 x-1\) (b) Explain why the two apparently different answers obtained in part (a) are really equivalent.

A traffic engineer monitors the rate at which cars enter the main highway during the afternoon rush hour. From her data she estimates that between 4: 30 PM. and 5: 30 P.M. the rate \(R(t)\) at which cars enter the highway is given by the formula \(R(t)=100\left(1-0.0001 t^{2}\right)\) cars per minute, where \(t\) is the time (in minutes) since 4: 30 PM. (a) When does the peak traffic flow into the highway occur? (b) Estimate the number of cars that enter the highway during the rush hour.

Use a calculating utility to find the midpoint approximation of the integral using \(n=20\) sub-intervals, and then find the exact value of the integral using Part 1 of the Fundamental Theorem of Calculus. $$\int_{1}^{3} \frac{1}{x^{2}} d x$$

Solve the initial-value problems. $$\frac{d y}{d t}=-e^{2 t}, y(0)=6$$

$$\begin{aligned} \sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=&\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right] \\\ &+\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right] \\ =& 5^{3}-1^{3}=124 \end{aligned}$$ For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises. $$\sum_{k=1}^{50}\left(\frac{1}{k}-\frac{1}{k+1}\right)$$
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