Problem 20 Evaluate using a substitution. (... [FREE SOLUTION]

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Calculus for Life Sciences

Marvin L. Bittinger, Neal Brand, John Quintanill

Biology

1 Edition

Chapter 5: Problem 20

Evaluate using a substitution. (Be sure to check by differentiating!) $$ \int \frac{d x}{1-x} $$

Short Answer

Expert verified

The integral $ \ \int \frac{dx}{1-x} \ $ is $ -\text{ln} |1-x| + C $.

Step by step solution

- Identify a suitable substitution

Select a substitution to simplify the integral. Let鈥檚 choose: \[ u = 1 - x \]

- Differentiate the substitution

Differentiate the chosen substitution with respect to $ x $: \[ du = -dx \] This gives: \[ dx = -du \]

- Rewrite the integral in terms of $ u $

Substitute $ u = 1 - x $ and $ dx = -du $ into the integral: \[ \begin{aligned} \ \ \ \int \frac{dx}{1-x} &= \int \frac{-du}{u} \ \ &= - \int \frac{du}{u} \ \ &= -\text{ln} |u| + C \ \ &= -\text{ln} |1-x| + C \ \ \ \end{aligned} \] where $ C $ is the constant of integration.

- Verify by differentiating the result

Differentiate $-\text{ln} |1-x| + C$ with respect to $ x $ to ensure it matches the original integrand: \[ \frac{d}{dx} (-\text{ln} |1-x| + C) = \frac{1}{1-x} \ \ \ \ \ \] which matches the original integrand, confirming that the substitution is correct.

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Key Concepts

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

Understanding Definite Integrals

A definite integral calculates the area under a curve between two given points. The limits of integration, often denoted as $a$ and $b$, specify these points. When computing a definite integral, you evaluate the antiderivative of the function at the upper limit and subtract the evaluation at the lower limit.

Definite integrals have the form: $\text{鈭珆_{a}^{b} f(x) \, dx$.

Some key properties of definite integrals include:

Linearity: you can split and combine integrals over the same interval.
Reversing limits flips the sign of the integral.
If the function is zero over an interval, the integral is also zero.

Using definite integrals, you can find areas, volumes, and other physical quantities.

Mastering Indefinite Integrals

Indefinite integrals, unlike definite integrals, do not have specific limits. Instead, they represent a family of functions whose derivative is the original function. When you compute an indefinite integral, you add a constant of integration, denoted as $C$, to account for all possible antiderivatives.

The general form of an indefinite integral is: $\text{鈭珆 f(x) \, dx = F(x) + C$.

This is because differentiation of the antiderivative and any constant yields the original function: \[ \frac{d}{dx} [F(x) + C] = f(x) \]

In practice, indefinite integrals help solve differential equations and find functions given their rates of change.

Fundamentals of Differentiation

Differentiation is a core concept in calculus that deals with finding the rate at which a function changes. The derivative of a function, denoted as $f'(x)$ or $\frac{d}{dx} f(x)$, measures the slope of the function at any point.

Basic rules of differentiation include:

Power rule: $ \frac{d}{dx} x^n = nx^{n-1} $
Product rule: $ \frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x) $
Quotient rule: $ \frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{u'(x) v(x) - u(x) v'(x)}{[v(x)]^2} $
Chain rule: $ \frac{d}{dx} f(g(x)) = f'(g(x)) g'(x) $

Utilizing these rules allows you to differentiate a wide variety of functions, essential for solving many problems in calculus.

Understanding U-Substitution in Integrals

The u-substitution method simplifies integrating complex functions. It involves substituting a part of the integrand with a new variable $u$ to make the integral more manageable.

The steps for u-substitution are:

Select a substitution: choose $u = g(x)$.
Differentiate $u$ to find $du = g'(x) \,dx$.
Rewrite the integral in terms of $u$.
Integrate with respect to $u$.
Substitute back the original variable.

As an example, consider the integral $ \text{鈭珆 \frac{dx}{1-x} $. We choose $ u = 1 - x $, differntiate to get $ du = -dx $, and substitute into the integral. This changes the integral to $ -\text{鈭珆 \frac{du}{u} = -\text{ln}|u| + C $, and substituting $u$ back gives the final answer: $-\text{ln}|1-x| + C$.

This example demonstrates how u-substitution simplifies the integral and helps find the antiderivative efficiently.

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91影视

Evaluate using a substitution. (Be sure to check by differentiating!) $$ \int \frac{d x}{1-x} $$

Short Answer

Step by step solution

- Identify a suitable substitution

- Differentiate the substitution

- Rewrite the integral in terms of \( u \)

- Verify by differentiating the result

Key Concepts

Understanding Definite Integrals

Mastering Indefinite Integrals

Fundamentals of Differentiation

Understanding U-Substitution in Integrals

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