Problem 20
Question
If the area bounded by the curve \(y=f(x), x\)-axis and the ordinates \(x=1\) and \(x=b\) is \((b-1) \sin (3 b+4)\), then (A) \(f(x)=\cos (3 x+4)+3(x-1) \sin (3 x+4)\) (B) \(f(x)=\sin (3 x+4)+3(x-1) \cos (3 x+4)\) (C) \(f(x)=\sin (3 x+4)-3(x-1) \cos (3 x+4)\) (D) None of these
Step-by-Step Solution
Verified Answer
The correct answer is (B): \(f(x)=\sin(3x+4)+3(x-1)\cos(3x+4)\).
1Step 1: Understand the Problem
We need to determine the function \(f(x)\) such that the area between the curve \(y=f(x)\), the \(x\)-axis, and the lines \(x=1\) and \(x=b\) equals \((b-1)\sin(3b+4)\).
2Step 2: Express Area as an Integral
The area under \(f(x)\) from \(x=1\) to \(x=b\) is given by the integral \(\int_{1}^{b} f(x) \, dx\). This is stated to be equal to \((b-1)\sin(3b+4)\).
3Step 3: Differentiate Both Sides
To find \(f(x)\), differentiate both sides with respect to \(b\). For the left-hand side, the derivative of \(\int_{1}^{b} f(x) \, dx\) with respect to \(b\) is \(f(b)\) by the Fundamental Theorem of Calculus. For the right-hand side, the derivative using the product rule is \(\sin(3b+4) + (b-1)\cdot 3\cos(3b+4)\).
4Step 4: Equate Derivatives
From differentiating both sides, we obtain:\[f(b) = \sin(3b+4) + 3(b-1)\cos(3b+4)\]
5Step 5: Adapt to Function Notation
Since we used \(b\) to denote the upper integration limit, we replace \(b\) with \(x\) to express \(f(x)\) in general terms:\[f(x) = \sin(3x+4) + 3(x-1)\cos(3x+4)\]
6Step 6: Match with Options
Compare the derived expression with the given options. We find that the expression matches Option B:\[f(x)=\sin(3x+4)+3(x-1)\cos(3x+4)\]
Key Concepts
Fundamental Theorem of CalculusArea under a curveDifferentiation
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a critical concept in Integral Calculus that bridges the processes of differentiation and integration. It consists of two main parts. The first part explains how the integral of a function is affected by its antiderivative. The second part states that the derivative of the integral of a function returns the original function itself.
In simple terms, this theorem demonstrates that differentiation and integration are inverse operations. In our exercise, when we differentiate the integral of the function from 1 to \(b\), the Fundamental Theorem shows us that we simply retrieve the function \(f(b)\).
This connection streamlines the process of calculating areas under curves because you can use antiderivatives rather than counting all the possible tiny areas individually. When you integrate a function from one point to another and then differentiate, you end up right back where you started— with the same function, as seen in our solution.
In simple terms, this theorem demonstrates that differentiation and integration are inverse operations. In our exercise, when we differentiate the integral of the function from 1 to \(b\), the Fundamental Theorem shows us that we simply retrieve the function \(f(b)\).
This connection streamlines the process of calculating areas under curves because you can use antiderivatives rather than counting all the possible tiny areas individually. When you integrate a function from one point to another and then differentiate, you end up right back where you started— with the same function, as seen in our solution.
Area under a curve
The concept of finding the 'area under a curve' refers to the process of using integration to calculate the space between the graph of a function and the \(x\)-axis. This space can represent a variety of meaningful quantities, such as accumulated values or total quantities over a range.
In our specific exercise, we are finding the area bounded by the curve \(y = f(x)\), the \(x\)-axis, and the lines \(x = 1\) and \(x = b\). The given area is specified as \((b-1)\sin(3b+4)\), which is a specific value that depends on the upper limit \(b\).
To find this area, we use the definite integral from 1 to \(b\), \(\int_{1}^{b} f(x) \, dx\). The process ensures that we compute the accumulated value of the function over the specified interval, giving us the enclosed area. This calculation is fundamental to solving many real-world problems where accumulation needs to be measured.
In our specific exercise, we are finding the area bounded by the curve \(y = f(x)\), the \(x\)-axis, and the lines \(x = 1\) and \(x = b\). The given area is specified as \((b-1)\sin(3b+4)\), which is a specific value that depends on the upper limit \(b\).
To find this area, we use the definite integral from 1 to \(b\), \(\int_{1}^{b} f(x) \, dx\). The process ensures that we compute the accumulated value of the function over the specified interval, giving us the enclosed area. This calculation is fundamental to solving many real-world problems where accumulation needs to be measured.
Differentiation
Differentiation is a process in calculus where we find the rate at which a function is changing at any given point. It provides a way to determine instantaneous rates of change, which is critical in fields like physics, economics, and engineering.
In the exercise, we used differentiation on both sides of the equation to derive the function \(f(x)\). By differentiating the integral \(\int_{1}^{b} f(x) \, dx\), and then \((b-1)\sin(3b+4)\), we utilize the principles of calculus to simplify and solve our problem efficiently.
Differentiating the integral involved using the Fundamental Theorem of Calculus, which led us to directly find \(f(b)\). For the right-hand side, applying the product rule gave us \(\sin(3b+4) + 3(b-1)\cos(3b+4)\). This calculated the rate of change of the given area expression, illustrating how differentiation can nicely complement integration.
In the exercise, we used differentiation on both sides of the equation to derive the function \(f(x)\). By differentiating the integral \(\int_{1}^{b} f(x) \, dx\), and then \((b-1)\sin(3b+4)\), we utilize the principles of calculus to simplify and solve our problem efficiently.
Differentiating the integral involved using the Fundamental Theorem of Calculus, which led us to directly find \(f(b)\). For the right-hand side, applying the product rule gave us \(\sin(3b+4) + 3(b-1)\cos(3b+4)\). This calculated the rate of change of the given area expression, illustrating how differentiation can nicely complement integration.
Other exercises in this chapter
Problem 18
If \([x]\) denotes the greatest integer \(\leq x\), then the value of \(\int_{4}^{10} \frac{\left[x^{2}\right]}{\left[x^{2}-28 x+196\right]+\left[x^{2}\right]}
View solution Problem 19
The area bounded by the \(y=|\sin x|, x\)-axis and the lines \(|x|=\pi\) is (A) 2 (B) 1 (C) 4 (D) None of these
View solution Problem 21
The area bounded by the curve \(y=\sin ^{-1} x\) and the lines \(x=0,|y|=\frac{\pi}{2}\) is (A) 2 (B) 4 (C) 8 (D) 16
View solution Problem 23
If \(\int_{0}^{100} f(x) d x=a\), then \(\sum_{r=1}^{100}\left(\int_{0}^{1} f(r-1+x) d x\right)=\) (A) \(100 a\) (B) \(a\) (C) 0 (D) \(100 a\)
View solution