/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 53

Evaluate the following integrals or state that they diverge. $$\int_{0}^{1} \ln x d x$$

Short Answer

Expert verified
Question: Evaluate the definite integral of the natural logarithm function, ln(x), on the interval from 0 to 1. Answer: The value of the integral is -1.

Step by step solution

01

Set up the integration by parts formula

To begin, recall the integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ We will choose our functions u and v to transform the given integral into this format.
02

Choose u and dv

Let's define: $$u = \ln x \quad \Rightarrow \quad \frac{du}{dx} = \frac{1}{x} \, dx \Rightarrow du = \frac{1}{x} \, dx$$ And, $$dv=dx \quad \Rightarrow \quad v=\int{1} dx \quad \Rightarrow \quad v=x$$
03

Apply integration by parts

Now, we apply the integration by parts formula to our problem: $$\int_{0}^{1} \ln x \, dx = \int u \, dv = uv - \int v \, du = \left[ x \ln x - \int x \cdot \frac{1}{x} dx \right]_0^1$$
04

Simplify and evaluate the integral

We can now simplify and evaluate: $$= \left[ x \ln x - \int 1 \, dx \right]_0^1 = \left[ x \ln x - x \right]_0^1 = \left[(1 \cdot \ln 1 - 1) - (0 \cdot \ln 0 - 0)\right]$$ Since \(\ln 1 = 0\), this simplifies to: $$= -1$$ So, the value of the integral is: $$\int_{0}^{1} \ln x \, dx = -1$$

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Key Concepts

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

Integration by Parts
Integration by parts is a powerful technique in calculus used to integrate the product of two functions. It's especially useful when dealing with integrals involving logarithmic, exponential, or trigonometric functions. The formula for integration by parts is: \[\int u \, dv = uv - \int v \, du\] Here鈥檚 how to use this method:
  • Choose: Select your functions \(u\) and \(dv\). A good choice usually simplifies the integral \(\int v \, du\).
  • Differentiate: Find \(du\) by differentiating \(u\), and integrate \(dv\) to find \(v\).
  • Substitute: Plug these into the integration by parts formula.
  • Simplify: Calculate the resulting expression, which often results in an easier integral.
In the example given, we set \(u = \ln x\) and \(dv = dx\), leading to \(du = \frac{1}{x} \, dx\) and \(v = x\). This substitution transforms the integral into a format where further simplification becomes possible.
Definite Integrals
Definite integrals calculate the net area under a curve within a specified range, from a lower limit \(a\) to an upper limit \(b\). The result of a definite integral is a number, representing this net area.To evaluate a definite integral:
  • Calculate: Perform the indefinite integration followed by applying both the upper and lower bounds.
  • Subtract: The formula is \(\left[ F(x) \right]_a^b = F(b) - F(a)\), where \(F(x)\) is the antiderivative of the integrand.
In the provided solution, the original integral \(\int_0^1 \ln x \, dx\) was evaluated using integration by parts. After simplifying, we find the value of the definite integral: \[\left[ x \ln x - x \right]_0^1\]Evaluating this at the bounds, the integral yields \(-1\). This value signifies the area below the curve \(\ln x\) from 0 to 1 but above the x-axis, considering orientation with respect to the axis.
Natural Logarithm
The natural logarithm is a logarithm to the base \(e\), where \(e\) is approximately 2.71828. The notation \(\ln x\) is used to represent the natural logarithm of \(x\).Properties of the natural logarithm include:
  • \(\ln 1 = 0\): Since any number to the power of zero equals one.
  • \(\ln e = 1\): By definition of \(e\).
When integrating functions involving \(\ln x\), integration by parts often comes in handy. Since differentiating \(u = \ln x\) is straightforward, this makes it a suitable candidate for \(u\) when using integration by parts. Understanding these properties aids in solving calculus problems involving natural logarithms, whether it's finding the derivative or evaluating integrals where \(\ln x\) is a factor.

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Most popular questions from this chapter

Evaluate the following integrals. $$\int \frac{68}{e^{2 x}+2 e^{x}+17} d x$$

Work Let \(R\) be the region in the first quadrant bounded by the curve \(y=\sqrt{x^{4}-4},\) and the lines \(y=0\) and \(y=2 .\) Suppose a tank that is full of water has the shape of a solid of revolution obtained by revolving region \(R\) about the \(y\) -axis. How much work is required to pump all the water to the top of the tank? Assume \(x\) and \(y\) are in meters.

Evaluate the following integrals. $$\int_{1}^{3} \frac{\tan ^{-1} \sqrt{x}}{x^{1 / 2}+x^{3 / 2}} d x$$

Preliminary steps The following integrals require a preliminary step such as a change of variables before using the method of partial fractions. Evaluate these integrals. $$\int \frac{\left(e^{3 x}+e^{2 x}+e^{x}\right)}{\left(e^{2 x}+1\right)^{2}} d x$$

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational function using the substitution \(u=\tan (x / 2)\) or, equivalently, \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ Verify relation \(A\) by differentiating \(x=2 \tan ^{-1} u .\) Verify relations \(B\) and \(C\) using a right-triangle diagram and the double-angle formulas $$\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} \quad \text { and } \quad \cos x=2 \cos ^{2} \frac{x}{2}-1$$.
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