Problem 125
Question
Suppose \(\int_{\pi}^{x} f(t) d t=\sin x+C\) for some constant \(C .\) Find the function \(f(x)\) and the constant \(C\).
Step-by-Step Solution
Verified Answer
The function \( f(x) \) is \( \cos x \), and the constant \( C \) is 0.
1Step 1: Introduce the Function
We are given the expression \( \int_{\pi}^{x} f(t) \, dt = \sin x + C \). This indicates that the integral of the function \( f(t) \) from \( t = \pi \) to \( t = x \) equals \( \sin x + C \). To find \( f(x) \), we need to recall that the derivative of an integral with a variable upper limit is the integrand evaluated at that point.
2Step 2: Differentiate the Expression
We apply the Fundamental Theorem of Calculus, which implies that if \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \). Differentiating the equation \( \int_{\pi}^{x} f(t) \, dt = \sin x + C \) with respect to \( x \):\[\frac{d}{dx}(\int_{\pi}^{x} f(t) \, dt) = \frac{d}{dx}(\sin x + C)\]This simplifies to:\[f(x) = \cos x\]
3Step 3: Determine the Constant
To find the constant \( C \), consider when \( x = \pi \). We know:\[\int_{\pi}^{\pi} f(t) \, dt = \sin(\pi) + C = 0 + C\]Since the integral from \( \pi \) to \( \pi \) results in 0, and \( \sin(\pi) = 0 \), it follows:\[0 = 0 + C \implies C = 0\]
4Step 4: Write the Function and Constant
Based on our calculations, the function \( f(x) \) is \( \cos x \), and the constant \( C = 0 \). We have found both the function for the integrand and the constant.
Key Concepts
Derivatives and IntegralsDefinite IntegralsTrigonometric Functions
Derivatives and Integrals
To solve the given problem, understanding the relationship between derivatives and integrals is key. Derivatives and integrals are fundamental concepts in calculus, operating as inverse processes. The problem involves the expression \( \int_{\pi}^{x} f(t) \, dt = \sin x + C \), which relates to the concept of finding the derivative and integral of a function.
To begin, it's vital to differentiate the given integral. This is where the derivative of an integral, as described by the Fundamental Theorem of Calculus, comes into play. The theorem states that if \( F(x) = \int_{a}^{x} f(t) \, dt \), then the derivative \( F'(x) \) is simply the function \( f(x) \) evaluated at \( x \). In simpler terms, taking the derivative of an integral with respect to its upper limit returns the original function inside the integral.
In the exercise, applying the derivative to both sides of the expression gives us \( f(x) = \cos x \). This effectively shows us that the function we seek is the cosine, a direct use of the relationship between derivatives and integrals.
To begin, it's vital to differentiate the given integral. This is where the derivative of an integral, as described by the Fundamental Theorem of Calculus, comes into play. The theorem states that if \( F(x) = \int_{a}^{x} f(t) \, dt \), then the derivative \( F'(x) \) is simply the function \( f(x) \) evaluated at \( x \). In simpler terms, taking the derivative of an integral with respect to its upper limit returns the original function inside the integral.
In the exercise, applying the derivative to both sides of the expression gives us \( f(x) = \cos x \). This effectively shows us that the function we seek is the cosine, a direct use of the relationship between derivatives and integrals.
Definite Integrals
Definite integrals have well-defined bounds, like the limits from \( a \) to \( b \), and they calculate the net area under a curve. In our exercise, we have \( \int_{\pi}^{x} f(t) \, dt \), a definite integral with the lower limit of \( \pi \) and the upper limit as \( x \).
The role of a definite integral in calculus is nuanced. It includes both concepts of area under a curve and accumulation of quantities. When evaluating definite integrals, special consideration is given to these limits.
The interesting part about definite integrals and variable limits, such as \( x \) in our case, is that they allow dynamic computation, adjusting as \( x \) changes. For the specific problem at hand, it results in expressing a dependence on the upper variable \( x \) through the function \( \sin x + C \). Here, applying the bounds to determine \( C \), we used the property that integral from a point to itself is zero: \( \int_{\pi}^{\pi} f(t) \, dt = 0 \). Given \( \sin(\pi) = 0 \), this yields \( C = 0 \).
The role of a definite integral in calculus is nuanced. It includes both concepts of area under a curve and accumulation of quantities. When evaluating definite integrals, special consideration is given to these limits.
The interesting part about definite integrals and variable limits, such as \( x \) in our case, is that they allow dynamic computation, adjusting as \( x \) changes. For the specific problem at hand, it results in expressing a dependence on the upper variable \( x \) through the function \( \sin x + C \). Here, applying the bounds to determine \( C \), we used the property that integral from a point to itself is zero: \( \int_{\pi}^{\pi} f(t) \, dt = 0 \). Given \( \sin(\pi) = 0 \), this yields \( C = 0 \).
Trigonometric Functions
Trigonometric functions like sine and cosine are fundamental in calculus, especially when dealing with periodic phenomena and integrals that arise from them. Our solution involved recognizing that the derivative of sine, \( \sin x \), is cosine, \( \cos x \).
Cosine, a primary trigonometric function, has a direct geometric interpretation on the unit circle and is periodic with a cycle of \( 2\pi \). Understanding these cyclical properties is crucial when integrating and differentiating trigonometric functions.
Within the context of the exercise, knowing that \( f(x) = \cos x \) illustrates a clear interplay between trigonometric identities and calculus. We take advantage of these identities to transition seamlessly between different forms, enabling us to accurately deduce functions from complex integrals.
Cosine, a primary trigonometric function, has a direct geometric interpretation on the unit circle and is periodic with a cycle of \( 2\pi \). Understanding these cyclical properties is crucial when integrating and differentiating trigonometric functions.
Within the context of the exercise, knowing that \( f(x) = \cos x \) illustrates a clear interplay between trigonometric identities and calculus. We take advantage of these identities to transition seamlessly between different forms, enabling us to accurately deduce functions from complex integrals.
- Sine and cosine are co-functions - meaning \( \cos(x) = \sin(\frac{\pi}{2} - x) \).
- The derivative relationship: \( \frac{d}{dx} \sin x = \cos x \) and \( \frac{d}{dx} \cos x = -\sin x \).
Other exercises in this chapter
Problem 123
Suppose that $$ \int_{0}^{x} f(t) d t=2 x^{2} $$ Find \(f(x)\).
View solution Problem 124
Suppose that $$ \int_{0}^{x} f(t) d t=e^{x}-1 $$ Find \(f(x)\).
View solution Problem 126
Suppose \(\int_{0}^{x} f(t) d t=\frac{1}{x^{2}+1}+C\) for some constant \(C\). Find the function \(f(x)\) and the constant \(C\).
View solution Problem 122
Evaluate the definite integrals. $$ \int_{1}^{2} \frac{2}{3 t+1} d t $$
View solution