Problem 20
Question
Find \(G^{\prime}(x)\).
$$
G(x)=\int_{1}^{x} \cos ^{3} 2 t \tan t d t ;-\pi / 2
Step-by-Step Solution
Verified Answer
\(G'(x) = \cos^3 (2x) \tan x\).
1Step 1: Identify the Problem
We are asked to find the derivative of the function \( G(x) = \int_{1}^{x} \cos^3 (2t) \tan t \, dt \). This is an application of the Leibniz's rule for differentiation with variable limits.
2Step 2: Apply the Fundamental Theorem of Calculus
According to the Fundamental Theorem of Calculus, if \( F(t) \) is an antiderivative of \( f(t) \), then \( \frac{d}{dx} \left( \int_{a}^{x} f(t) \, dt \right) = f(x) \). Here \( f(t) = \cos^3 (2t) \tan t \) and consequently, \( G'(x) = \cos^3 (2x) \tan x \).
3Step 3: Verify the Correctness of the Derivative
Given the function \( G(x) \) being defined with a variable upper limit of integration, it is crucial to apply differentiation correctly as per the Fundamental Theorem of Calculus. As verified, \( G'(x) = \cos^3 (2x) \tan x \).
Key Concepts
Leibniz's RuleIntegration with Variable LimitsDifferentiation of Integrals
Leibniz's Rule
Leibniz's rule is a critical tool in calculus for differentiating integrals that have variable limits. It is a generalization of the fundamental theorem of calculus. This principle proves particularly useful in solving problems where the limits of integration aren't constants but functions of a variable, such as 'x' in our exercise.
The core idea is that Leibniz's rule allows us to take the derivative of an integral whose limits are functions of x. In simple terms, if you have a function that is created by integrating from a variable lower limit a(x) to a variable upper limit b(x), Leibniz's rule gives us a way to find the derivative of this integral with respect to x.
The core idea is that Leibniz's rule allows us to take the derivative of an integral whose limits are functions of x. In simple terms, if you have a function that is created by integrating from a variable lower limit a(x) to a variable upper limit b(x), Leibniz's rule gives us a way to find the derivative of this integral with respect to x.
- The rule states that the derivative of an integral \( \int_{a(x)}^{b(x)} f(t, x) \, dt \) with respect to 'x' is a combination of:
- Evaluating the integrand at the upper limit times the derivative of the upper limit;
- Subtracting the integrand evaluated at the lower limit times the derivative of the lower limit;
- And adding the integral of the partial derivative of f with respect to x.
Integration with Variable Limits
Integration with variable limits is a fascinating aspect of calculus that takes us beyond the constant limit scenarios. Here, we're evaluating an integral where the endpoint isn't just a number but a variable, such as 'x', which changes the scope and output of the integral itself.
In the exercise, the function is defined as an integral from a constant lower limit 1 to a variable upper limit x. This denotes that as 'x' changes, so does the value of the integral. The integral thus acts like a function, where each 'x' corresponds to a different area under the curve of the function \( \cos^3(2t) \tan t \) from 1 to 'x'.
In the exercise, the function is defined as an integral from a constant lower limit 1 to a variable upper limit x. This denotes that as 'x' changes, so does the value of the integral. The integral thus acts like a function, where each 'x' corresponds to a different area under the curve of the function \( \cos^3(2t) \tan t \) from 1 to 'x'.
- When the limit is a variable, it affects how we manipulate the integral in operations such as differentiation.
- This becomes handy for finding how fast or slow the accumulated area changes as the upper boundary moves.
Differentiation of Integrals
Differentiating integrals is central to connecting antiderivatives and area functions. By using the fundamental theorem of calculus, it allows us to transition smoothly from integral calculus back to differential calculus, binding together these seemingly distinct branches.
When we differentiate an integral like \( G(x) = \int_{1}^{x} \cos^3(2t) \tan t \, dt \), we're essentially inquiring how the rate of change of the function, as expressed through the integral, behaves. The theorem simplifies this by relying on the understanding that if \( F(t) \) is an antiderivative of \( f(t) \), then:
In the exercise, this differentiation leads us directly to \( G'(x) = \cos^3(2x) \tan x \), illustrating how neatly integration and differentiation fit together. This relationship is crucial in calculus because it means we can not only compute areas but also understand the dynamics of changing states within a calculus function that have been set up as integrals.
When we differentiate an integral like \( G(x) = \int_{1}^{x} \cos^3(2t) \tan t \, dt \), we're essentially inquiring how the rate of change of the function, as expressed through the integral, behaves. The theorem simplifies this by relying on the understanding that if \( F(t) \) is an antiderivative of \( f(t) \), then:
- \( \frac{d}{dx} \left( \int_{a}^{x} f(t) \, dt \right) = f(x) \)
In the exercise, this differentiation leads us directly to \( G'(x) = \cos^3(2x) \tan x \), illustrating how neatly integration and differentiation fit together. This relationship is crucial in calculus because it means we can not only compute areas but also understand the dynamics of changing states within a calculus function that have been set up as integrals.
Other exercises in this chapter
Problem 20
Use Special Sum Formulas 1-4 to find each sum. $$ \sum_{i=1}^{10}[(i-1)(4 i+3)] $$
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use the method of substitution to find each of the following indefinite integrals. $$ \int \cos (\pi v-\sqrt{7}) d v $$
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Calculate \(\int_{a}^{b} f(x) d x\), where a and \(b\) are the left and right end points for which f is defined, by using the Interval Additive Property and the
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