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What change of variables would you use for the integral \(\int(4-7 x)^{-6} d x ?\)

Short Answer

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Question: Evaluate the integral: $$\int(4-7x)^{-6}dx.$$ Answer: The integral can be evaluated as $$\int(4-7x)^{-6}dx = \frac{1}{35}(4-7x)^{-5} + C$$.

Step by step solution

01

Identify a possible substitution

Let's identify a good substitution for the integral: $$\int(4-7x)^{-6}dx.$$ Notice that \((4-7x)^{-6}\) is the major component of the integral. So we can consider the substitution $$u = 4 - 7x.$$
02

Differentiate the substitution and isolate dx

Now we need to find the derivative of \(u\) with respect to \(x\) and isolate \(dx\): $$\frac{du}{dx} = -7$$ $$du = -7dx$$ $$dx = \frac{-1}{7}du$$
03

Make the substitution and evaluate the new integral

Replace our substitution \(u = 4 - 7x\) and \(dx = \frac{-1}{7}du\) in the integral: $$\int(4-7x)^{-6}dx = \int u^{-6} \left(\frac{-1}{7} du\right) = \frac{-1}{7}\int u^{-6}du$$
04

Evaluate the integral

Calculate the integral: $$\frac{-1}{7}\int u^{-6}du = \frac{-1}{7}\left[\frac{u^{-5}}{-5}\right] + C$$
05

Replace u back in terms of x

Finally, substitute \(u\) back in terms of \(x\): $$\frac{-1}{7}\left[\frac{(4-7x)^{-5}}{-5}\right] + C = \frac{1}{35}(4-7x)^{-5} + C$$ The resulting integral after the change of variables using the substitution method is: $$\int(4-7x)^{-6}dx = \frac{1}{35}(4-7x)^{-5} + C$$

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Key Concepts

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

Change of Variables
The technique known as "change of variables" is fundamental in integral calculus for simplifying complex integrals. Essentially, it involves replacing a variable with another variable to make integration more manageable. This method is helpful when the integrand (the function you're integrating) is complicated. By strategically choosing a new variable, you can transform the difficult portion of the integrand into a simpler form.
In the context of the substitution we examined, the expression \(4-7x\) was chosen as the part to substitute with a new variable \(u\). We then expressed it as \(u = 4 - 7x\). This substitution turns the complex function into a simpler function of \(u\), which is much easier to integrate.
Using this substitution effectively simplified both the integrand and the process of integrating. By changing variables, you can often transform integrals into more familiar or solvable forms. This is why it's a widely used method in calculus for handling integrals that are not straightforward.
Integral Calculus
Integral calculus is a crucial branch of mathematics focusing on the concept of integration. The fundamental aim of integral calculus is to find the total accumulation of quantities, like area under a curve or total change over an interval.
Integration is the reverse process of differentiation, making it a powerful tool in mathematics for solving problems related to areas, volumes, and other quantities.
  • Definite Integrals - Provide a number (representing the area under the curve bounded by the function, the x-axis, and the limits of integration).
  • Indefinite Integrals - Provide a family of functions and are solved inclusively with a constant of integration, usually denoted as \(C\).
Applying integration techniques like substitution allows us to solve complex problems, just as we solved the integral \(\int(4-7x)^{-6}dx\) by changing variables.
U-Substitution
U-substitution, also known as integration by substitution, is akin to the "chain rule" used in differentiation. It involves changing variables to simplify the integral.
  • Step 1: Choose a substitution. Pick a part of the integrand you want to replace, here it was \(u = 4 - 7x\).
  • Step 2: Find the derivative. Calculate \(\frac{du}{dx}\) and solve for \(dx\), becoming \(dx = \frac{-1}{7}du\).
  • Step 3: Substitute and transform the integral. Replace in the original integral and restructure it as \(\int u^{-6}\).
  • Step 4: Integrate and back-substitute. Integrate \(u\) and revert \(u\) back to \(x\) giving \(\frac{1}{35}(4-7x)^{-5} + C\).
The primary purpose of u-substitution is to make an integral more straightforward to analyze and solve. It transforms parts of the integrand step-by-step, enabling more accessible engagement with the integration process.

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Most popular questions from this chapter

A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ where we assume s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated using integration by parts: $$F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1}$$ Verify the following Laplace transforms, where a is a real number. $$f(t)=t \rightarrow F(s)=\frac{1}{s^{2}}$$

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A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ where we assume s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated using integration by parts: $$F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1}$$ Verify the following Laplace transforms, where a is a real number. $$f(t)=1 \rightarrow F(s)=\frac{1}{s}$$

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