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Chapter 5: Problem 68

Simplify the following expressions. $$\frac{d}{d x} \int_{e^{x}}^{e^{2 x}} \ln t^{2} d t$$

Short Answer

Expert verified
Question: Find the derivative of the given definite integral with variable limits: $$\frac{d}{dx} \int_{e^x}^{e^{2x}} \ln(t^2) dt$$ Answer: $$2e^x(4xe^x - 2e^x - 2x + 2)$$

Step by step solution

01

Find the antiderivative of the integrand

To find the antiderivative of \(\ln(t^2)\), we can use integration by parts with \(u = \ln(t^2)\) and \(dv = dt\). First, let's find the necessary components: $$du = \frac{2}{t} dt$$ $$v = t$$ Now, we can apply integration by parts: $$\int \ln(t^2) dt = t \ln(t^2) - \int 2 dt$$ $$\int \ln(t^2) dt = t \ln(t^2) - 2t + C$$
02

Apply Leibniz's rule

Using the antiderivative we found in Step 1, we can apply Leibniz's rule to the original expression. We have: $$\frac{d}{dx} \int_{e^x}^{e^{2x}} \ln(t^2) dt = \left[e^{2x} \ln(e^{4x}) - 2e^{2x} \right](2e^{2x}) - \left[e^x \ln(e^{2x}) - 2e^x \right](e^x)$$
03

Simplify the expression

Now we will simplify the obtained expression: $$\left[2e^{2x}(4x - 2) - e^x(2x - 2) \right]$$ Factor out the common terms: $$2e^{x}(4x - 2)e^x - 2e^x(2x - 2)$$ $$2e^x(e^x(4x-2) - (2x-2))$$ And we can further simplify the expression inside the parenthesis: $$2e^x(4xe^x - 2e^x - 2x + 2)$$ The final simplified expression is: $$\frac{d}{dx} \int_{e^x}^{e^{2x}} \ln(t^2) dt = 2e^x(4xe^x - 2e^x - 2x + 2)$$

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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 very useful technique in calculus for finding the integral of products of functions. It leverages the product rule for differentiation and works particularly well when you want to simplify an integral involving a product of two functions.
To perform integration by parts, you select one part of the product to differentiate and another to integrate. The formula for integration by parts is:
  • \( \int u \, dv = uv - \int v \, du \)
In applying this formula, follow these steps:
  • Choose \( u \) and \( dv \) from the integrand, where the choice often depends on the ease of finding the derivative and the antiderivative.
  • Differentiate \( u \) to get \( du \).
  • Integrate \( dv \) to get \( v \).
  • Substitute into the formula \( uv - \int v \, du \) and solve the integral.
Remember the rule "LIATE" to select \( u \), prioritizing Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, then Exponential functions.
In this exercise, we set \( u = \ln(t^2) \) and \( dv = dt \) to find the antiderivative needed in the solution.
Antiderivative
An antiderivative, also known as an indefinite integral, is a function whose derivative is the given function. If we have a function \( f(x) \), then an antiderivative \( F(x) \) satisfies \( F'(x) = f(x) \). Integrating \( f(x) \) without specified bounds of integration gives us this antiderivative.
In the context of the original problem, we sought to find the antiderivative of \( \ln(t^2) \). By applying integration by parts, we broke down the function into manageable parts and simplified the integration process.
Always remember that the antiderivative includes a constant of integration \( C \) since differentiation erases constant terms. This constant is considered when dealing with indefinite integrals, although it is not directly relevant in definite integrals as in our exercise.
Integral Calculus
Integral calculus is a branch of mathematics focused on the concept of integration. It is used to find the area under a curve, accumulate quantities, and solve differential equations. Integral calculus helps answer key questions about accumulation and total change.
There are two primary types of integrals in this field:
  • Indefinite Integrals: Deals with finding the antiderivative of a function, providing a general form without specific limits.
  • Definite Integrals: Computes the accumulation of a quantity, bounded between two specific limits, essentially resulting in a number.
In solving our given problem, we used a special technique known as Leibniz's Rule, which involves differentiating under the integral sign. This rule helped us cope with changing limits \( e^x \) to \( e^{2x} \), allowing us to differentiate an integral with variable limits effectively.
Integral calculus is essential for exploring accumulative relationships in science and engineering, making it a vital mathematical tool.

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Most popular questions from this chapter

Use a change of variables to find the following indefinite integrals. Check your work by differentiating. $$\int x e^{x^{2}} d x$$

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A function \(f\) is symmetric about a point \((c, d)\) if whenever \((c-x, d-y)\) is on the graph, then so is \((c+x, d+y) .\) Functions that are symmetric about a point \((c, d)\) are easily integrated on an interval with midpoint \(c\). a. Show that if \(f\) is symmetric about \((c, d)\) and \(a>0,\) then \(\int_{c-a}^{c+a} f(x) d x=2 a f(c)=2 a d\). b. Graph the function \(f(x)=\sin ^{2} x\) on the interval \([0, \pi / 2]\) and show that the function is symmetric about the point \(\left(\frac{\pi}{4}, \frac{1}{2}\right)\). c. Using only the graph of \(f\) (and no integration), show that \(\int_{0}^{\pi / 2} \sin ^{2} x d x=\frac{\pi}{4} .\) (See the Guided Project Symmetry in Integrals.)

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