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Chapter 13: Problem 2

Determine \(\int\left(6 . e^{3 x-5}+\frac{4}{3 x-2}-5^{2 x+1}\right) \mathrm{d} x\).

Short Answer

Expert verified
The integral of the given expression is the sum of the integrals of each separate term.

Step by step solution

01

Break Down the Integral

The integral given is \(\int\left(6 \cdot e^{3x-5}+\frac{4}{3x-2}-5^{2x+1}\right) \mathrm{d}x\). We need to separate this into three individual integrals for ease of integration: \(\int 6 \cdot e^{3x-5} \, \mathrm{d}x\), \(\int \frac{4}{3x-2} \, \mathrm{d}x\), and \(\int -5^{2x+1} \, \mathrm{d}x\).

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Key Concepts

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

Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. The most common base in calculus is the natural exponential, denoted as \(e\). This constant is approximately equal to 2.71828, and it is uniquely important because its derivative with respect to its exponent is itself: \(\frac{d}{dx}(e^x) = e^x\).

Exponential functions are found everywhere in calculus due to their unique properties. When dealing with exponentials, one must pay attention to the base and the exponent's form.
  • The base of the exponential function indicates the growth rate. Larger bases grow faster.
  • In integrals, exponential functions like \(e^{3x-5}\) become manageable once we recognize their structure and differentiate or integrate accordingly.
  • Using substitution can be especially effective when the exponent of the exponential has a variable term like \(3x-5\).
By breaking down exponential components in an integral, we simplify complex expressions into parts we can integrate separately.
Integration by Substitution
Integration by substitution is a technique used to simplify integrals, making them easier to evaluate. It is closely related to the chain rule in differentiation. The core idea is to transform the original variable into a new variable, simplifying the integral's expression.

This process involves a few steps:
  • Identify a part of the integral that can be set as a substitution. This is usually an inner expression where a variable change simplifies the integral.
  • Let \(u\) equal this inner expression. For example, if we have \(u = 3x - 5\), then \(\frac{du}{dx} = 3\), and thus \(dx = \frac{1}{3}du\).
  • Substitute \(u\) and \(dx\) into the integral, transforming it into \(\int f(u) \, du\).
  • Integrate with respect to \(u\), and then substitute back the original variables to get the final solution.
This method simplifies complex integrals, allowing easier computation while maintaining accuracy.
Logarithmic Integration
Logarithmic integration is used when the integrand is in the form \(\frac{1}{x}\) or more complex forms like \(\frac{4}{3x-2}\). Such integrals often lead to natural logarithms as solutions. This technique can be seen as the reverse process of differentiating logarithmic functions.

Consider the integral \(\int \frac{1}{x} \, dx = \ln|x| + C\). For more complex versions:
  • Identify parts of the integral that fit the pattern of a logarithmic derivative, such as \(\frac{4}{3x-2}\). This is a classic case for substitution.
  • Use a substitution, like letting \(u = 3x - 2\). This turns the integral into \(\frac{4}{u} \, du\).
  • Integrate to find \(4 \ln|u| + C\) and then substitute back \(u\) to \(4 \ln|3x - 2| + C\).
Logarithmic integration provides a straightforward way to tackle integrals that appear complicated, simplifying them into recognizable logarithmic forms.

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Most popular questions from this chapter

In each of the following, determine the area enclosed by the given boundaries: (a) \(y=10-x^{2}\) and \(y=x^{2}+2\) (b) \(y=x^{2}-4 x+20, \quad y=3 x, \quad x=0\) and \(x=4\) (c) \(y=4 e^{2 x}, \quad y=4 e^{-x}, \quad x=1\) and \(x=3\) (d) \(y=5 e^{x}, \quad y=x^{3}, \quad x=1\) and \(x=4\) (e) \(y=20+2 x-x^{2}, \quad y=\frac{e^{x}}{2}, \quad x=1\) and \(x=3\)

Determine the following: (a) \(\int(5-6 x)^{2} \mathrm{~d} x\) (e) \(\int \sqrt{5-3 x} \mathrm{~d} x\) (h) \(\int 4 \cos (1-2 x) d x\) (b) \(\int 4 \sin (3 x+2) d x\) (f) \(\int 6 \cdot e^{3 x+1} \mathrm{~d} x\) (i) \(\int 3 \sec ^{2}(4-3 x) d x\) (c) \(\int \frac{5}{2 x+3} d x\) (g) \(\int(4-3 x)^{-2} \mathrm{~d} x\) (j) \(\int 8 . e^{3-4 x} \mathrm{~d} x\) (d) \(\int 3^{2 x-1} \mathrm{~d} x\)

If \(I=\int\left(8 x^{3}+3 x^{2}-6 x+7\right) \mathrm{d} x\), determine the value of \(I\) when \(x=-3\) given that when \(x=2, I=50\)

Determine the following using partial fractions: (a) \(\int \frac{6 x+1}{4 x^{2}+4 x-3} d x\) (b) \(\int \frac{x+11}{x^{2}-3 x-4} d x\) (c) \(\int \frac{3 x-17}{12 x^{2}-19 x+4} d x\) (d) \(\int \frac{20 x+2}{8 x^{2}-14 x-15} d x\) (e) \(\int \frac{6 x^{2}-2 x-2}{6 x^{2}-7 x+2} d x\) (f) \(\int \frac{4 x^{2}-9 x-19}{2 x^{2}-7 x-4} d x\) (g) \(\int \frac{6 x^{2}+2 x+1}{2 x^{2}+x-6} d x\) (h) \(\int \frac{4 x^{2}-2 x-11}{2 x^{2}-7 x-4} d x\)

(a) Evaluate the area under the curve \(y=5 . e^{x}\) between \(x=0\) and \(x=3\). (b) Show that the straight line \(y=-2 x+28\) crosses the curve \(y=36-x^{2}\) at \(x=-2\) and \(x=4\). Hence, determine the area enclosed by the curve \(y=36-x^{2}\) and the line \(y=-2 x+28\)
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