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Chapter 6: Problem 34

Find the value of \(k\) that makes the antidifferentiation formula true. [Note: You can check your answer without looking in the answer section. How?] $$\int(2 x-1)^{3} d x=k(2 x-1)^{4}+C$$

Short Answer

Expert verified
The value of k is \( \frac{1}{8} \).

Step by step solution

01

- Recognize the Given Indefinite Integral

The given indefinite integral is \( \int (2x-1)^3 \, dx \). The goal is to find the value of \( k \) that makes the antidifferentiation formula true.
02

- Perform a Substitution

Let \( u = 2x - 1 \). Therefore, \( du = 2 \, dx \), or \( dx = \frac{du}{2} \). The integral becomes: \[ \int (2x-1)^3 \, dx = \int u^3 \cdot \frac{1}{2} \, du = \frac{1}{2} \int u^3 \, du \]
03

- Apply the Power Rule for Integration

The power rule states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). Applying this rule to our integral: \[ \frac{1}{2} \int u^3 \, du = \frac{1}{2} \left( \frac{u^{4}}{4} \right) + C = \frac{1}{8} u^4 + C \]
04

- Substitute Back the Original Variable

Recall that \( u = 2x - 1 \). Substitute \( u \) back into the expression: \[ \frac{1}{8} (2x - 1)^4 + C \]
05

- Identify \( k \)

Compare the resulting expression \( \frac{1}{8} (2x - 1)^4 + C \) with the form \( k(2x - 1)^4 + C \). It is evident that \( k = \frac{1}{8} \).
06

- Verify the Solution

Differentiate \( k(2x - 1)^4 + C \) to ensure it matches the original integrand: \[ \frac{d}{dx} \left( \frac{1}{8} (2x - 1)^4 + C \right) = \frac{1}{8} \cdot 4(2x - 1)^3 \cdot 2 = (2x - 1)^3 \]. This matches the original integrand.

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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 type of integral that calculates the net area under a curve within specified limits, usually between points \(a\) and \(b\).
Unlike an indefinite integral, which represents a family of functions and includes a constant \(C\), a definite integral gives a numerical value.
The notation for a definite integral is \[ \int_a^b f(x) \, dx. \]
Here鈥檚 a simple way to understand it:
  • Take the definite integral of a function to find the accumulated quantity over an interval.

  • For example, if the function represents velocity over time, the definite integral gives the total distance traveled.

  • The computation involves finding the antiderivative (indefinite integral) first and then evaluating this antiderivative at the boundary points \(a\) and \(b\).

So, when dealing with definite integrals, you're focusing on how much one variable accumulates within specific limits.
substitution method
The substitution method is a powerful tool in integration, especially for handling more complex integrals. It simplifies an integral by changing variables.
Here's how it works:
  • Identify a part of the integrand that can be set as a new variable, say \(u\).

  • Replace the chosen part with \(u\) and determine \(du\) (the differential of \(u\)).

  • Substitute both \(u\) and \(du\) into the integral, transforming it into an easier problem.
  • Integrate with respect to \(u\).
  • Finally, substitute back the original variable.

Let's see it in action through the example given: the original integral is: \( \int (2x - 1)^3 \, dx \).
Set \( u = 2x - 1 \). Thus, \( du = 2 \, dx \) or \( dx = \frac{du}{2} \). The integral changes to: \[ \int u^3 \cdot \frac{1}{2} \, du = \frac{1}{2} \int u^3 \, du. \]
This replacement makes the integral much more straightforward to compute.
power rule for integrals
The power rule for integrals helps in integrating functions of the form \( u^n \).
The rule states: \[ \int u^n \, du = \frac{u^{n+1}}{n+1} + C, \] where \( n \) is any real number except \( -1 \).
Let's break it down:
  • It applies when the integrand is a power of the variable, making the integration process direct.

  • Increment the exponent by 1 to get \( n+1 \).

  • Divide by this new exponent.

  • Don鈥檛 forget to add the constant of integration \( C \).

Applying the rule to the transformed integral from the example: \[ \frac{1}{2} \int u^3 \, du. \]
Using the power rule, integrate \( u^3 \) as follows: \[ \frac{1}{2} \cdot \frac{u^4}{4} + C = \frac{1}{8} u^4 + C. \] This shows just how handy the power rule is.
Integral calculus becomes much more manageable with these key rules!

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Most popular questions from this chapter

Evaluate a Riemann sum to approximate the area under the graph of \(f(x)\) on the given interval, with points selected as specified. \(f(x)=x \sqrt{1+x^{2}} ; 1 \leq x \leq 3, n=20,\) midpoints of subintervals.

We show that, as the number of subintervals increases indefinitely, the Riemann sum approximation of the area under the graph of \(f(x)=x^{2}\) from 0 to 1 approaches the value \(\frac{1}{3},\) which is the exact value of the area. Verify the given formula for \(n=1,2,3,4\) $$1^{2}+2^{2}+\cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$$

The velocity of a skydiver at time \(t\) seconds is \(v(t)=45-45 e^{-0.2 t}\) meters per second. Find the distance traveled by the skydiver the first 9 seconds.

A rock is dropped from the top of a 400 -foot cliff. Its velocity at time \(t\) seconds is \(v(t)=-32 t\) feet per second. (a) Find \(s(t),\) the height of the rock above the ground at time \(t\) (b) How long will the rock take to reach the ground? (c) What will be its velocity when it hits the ground?

Find all functions \(f(t)\) that satisfy the given condition. $$f^{\prime}(x)=\sqrt{x}+1, f(4)=0$$
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