Problem 7
Question
Evaluate the following definite integrals. If \(f^{\prime}(x)=g(x)\) and \(g\) is a continuous function for all real values of \(x\) express \(\int_{1}^{2} g(4 x) d x\) in terms of \(f\).
Step-by-Step Solution
Verified Answer
Question: Evaluate the definite integral \(\int_{1}^{2} g(4x) dx\) and express the result in terms of \(f\), given that \(f^{\prime}(x)=g(x)\).
Answer: \(\int_{1}^{2} g(4x) dx = \frac{1}{4} (f(8) - f(4))\).
1Step 1: Perform a substitution
Let \(u = 4x\). Then, \(\frac{du}{dx} = 4\), or \(dx = \frac{1}{4} du\). Now, we need to change the limits of integration as well. When \(x=1\), we have \(u=4(1)=4\). Similarly, when \(x=2\), we have \(u=4(2)=8\). So, the integral becomes:
\(\int_{1}^{2} g(4x) dx = \int_{4}^{8} g(u) \frac{1}{4} du\)
2Step 2: Express the integral in terms of \(f\)
We are given that \(f^{\prime}(x)=g(x)\). Using this information, we have \(f^{\prime}(u) = g(u)\). Therefore, the integral can be rewritten as:
\(\int_{4}^{8} f^{\prime}(u) \frac{1}{4} du\)
3Step 3: Apply the Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus informs us that if \(F(x)\) is an antiderivative of a continuous function \(f(x)\) on the interval \([a, b]\), then \(\int_{a}^{b} f(x) dx = F(b) - F(a)\).
Applying the theorem to our integral, we obtain:
\(\int_{4}^{8} f^{\prime}(u) \frac{1}{4} du = \frac{1}{4} (f(8) - f(4))\)
Hence, the definite integral \(\int_{1}^{2} g(4x) dx\) can be expressed in terms of \(f\) as:
\(\boxed{\int_{1}^{2} g(4x) dx = \frac{1}{4} (f(8) - f(4))}\).
Key Concepts
U-SubstitutionFundamental Theorem of CalculusAntiderivativeAP Calculus AB
U-Substitution
U-substitution is a powerful technique used in calculus to simplify the process of integrating complex functions. It involves replacing a part of the function with a new variable, typically denoted as 'u', which simplifies the integral into a more manageable form. Think of it like performing a substitution in algebra to make an equation easier to solve.
For example, in the exercise given, we have the integral \(\int_{1}^{2} g(4x) dx\) which looks a bit complicated due to the presence of \(4x\) inside the function \(g\). By setting \(u = 4x\), we are transforming the integral into a new variable \(u\) which makes \(g(4x)\) become \(g(u)\), and \(dx\) becomes \(\frac{1}{4} du\) due to the derivative of \(u\) with respect to \(x\) being \(4\). This substitution significantly simplifies the integral, allowing us to use standard integration techniques.
For example, in the exercise given, we have the integral \(\int_{1}^{2} g(4x) dx\) which looks a bit complicated due to the presence of \(4x\) inside the function \(g\). By setting \(u = 4x\), we are transforming the integral into a new variable \(u\) which makes \(g(4x)\) become \(g(u)\), and \(dx\) becomes \(\frac{1}{4} du\) due to the derivative of \(u\) with respect to \(x\) being \(4\). This substitution significantly simplifies the integral, allowing us to use standard integration techniques.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a central principle that connects the concept of a derivative with that of an integral. It asserts that if \(F(x)\) is the antiderivative of a continuous function \(f(x)\), then the definite integral of \(f(x)\) over the interval \( [a, b] \) can be evaluated by computing \(F(b) - F(a)\).
In simpler terms, the theorem tells us that to find the area under the curve of \(f(x)\) from \(a\) to \(b\), we can simply take the difference between the values of one antiderivative of \(f(x)\) at these two endpoints. This provides a straightforward method to solve definite integrals, as seen in the provided exercise solution. Once we've integrated the function with respect to \(u\), we apply the theorem by substituting the new limits back into the antiderivative, finding \(\frac{1}{4} (f(8) - f(4))\) as the result.
In simpler terms, the theorem tells us that to find the area under the curve of \(f(x)\) from \(a\) to \(b\), we can simply take the difference between the values of one antiderivative of \(f(x)\) at these two endpoints. This provides a straightforward method to solve definite integrals, as seen in the provided exercise solution. Once we've integrated the function with respect to \(u\), we apply the theorem by substituting the new limits back into the antiderivative, finding \(\frac{1}{4} (f(8) - f(4))\) as the result.
Antiderivative
An antiderivative, also commonly referred to as an indefinite integral, is a function \(F(x)\) that represents the accumulation of the area under the curve of a given function \(f(x)\). It is essentially a reverse process of differentiation: if the derivative of \(F(x)\) with respect to \(x\) is \(f(x)\), written as \(F'(x) = f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\).
It is important to note that because the derivative of a constant is zero, the antiderivative of a function is not unique; it actually encompasses a family of functions, all differing by a constant. This is why when evaluating definite integrals using antiderivatives, as with our exercise, we don’t need to worry about the constant since it cancels out when we subtract the antiderivative values at the upper and lower limits.
It is important to note that because the derivative of a constant is zero, the antiderivative of a function is not unique; it actually encompasses a family of functions, all differing by a constant. This is why when evaluating definite integrals using antiderivatives, as with our exercise, we don’t need to worry about the constant since it cancels out when we subtract the antiderivative values at the upper and lower limits.
AP Calculus AB
AP Calculus AB is an advanced placement course and exam offered by the College Board in the United States. It covers differential and integral calculus at a level that is equivalent to a first-semester college calculus course. Topics include limits, derivatives, definite integrals, the Fundamental Theorem of Calculus, and their applications.
Understanding concepts like u-substitution and the Fundamental Theorem of Calculus is crucial for success in AP Calculus AB. Exercises similar to the one provided reinforce the practical applications of these concepts, guiding students towards a deeper comprehension of calculus and helping them prepare for the AP exam. The ability to express a definite integral in terms of an antiderivative, as shown in the provided exercise, is a fundamental skill that students are expected to demonstrate in the AP Calculus AB exam.
Understanding concepts like u-substitution and the Fundamental Theorem of Calculus is crucial for success in AP Calculus AB. Exercises similar to the one provided reinforce the practical applications of these concepts, guiding students towards a deeper comprehension of calculus and helping them prepare for the AP exam. The ability to express a definite integral in terms of an antiderivative, as shown in the provided exercise, is a fundamental skill that students are expected to demonstrate in the AP Calculus AB exam.
Other exercises in this chapter
Problem 5
Evaluate the following definite integrals. If \(\int_{0}^{k}(6 x-1) d x=4,\) find \(k\).
View solution Problem 6
Evaluate the following definite integrals. $$ \int_{0}^{\pi} \frac{\sin x}{\sqrt{1+\cos x}} d x $$
View solution Problem 8
Evaluate the following definite integrals. $$ \int_{\ln 2}^{\ln 3} 10 e^{x} d x $$
View solution Problem 9
Evaluate the following definite integrals. $$ \int_{e}^{e^{2}} \frac{1}{t+3} d t $$
View solution