Question

If \(g(x)\) is the inverse of \(f(x)\) and \(f(x)\) has domain \(x \in[1,5]\), where \(f(1)=2\) and \(f(5)=10\) then find the value of \(\int_{1}^{5} f(x) d x+\int_{2}^{10} g(y) d y\).

Short answer

Question: Calculate the value of the expression, given \(f(x)\) and its inverse function \(g(y)\) with the following properties: \(f(1)=2\), \(f(5)=10\), and \(x \in [1,5]\). Answer: The value of the expression \(\int_{1}^{5} f(x) dx + \int_{2}^{10} g(y) dy\) can be simplified to \(\int_{1}^{5} f(x) dx + 48\). Without the explicit function \(f(x)\), we cannot find the exact value for \(\int_{1}^{5} f(x) dx\).

Step-by-step solution

Relationship between a function and its inverse

A function and its inverse "undo" each other. For instance, if \(g(y)\) is the inverse of \(f(x)\), then: 1. \(g(f(x)) = x\) 2. \(f(g(y)) = y\) Also, note that the graph of function and its inverse are reflections about the line \(y=x\).

Apply given conditions

We start with the domain of \(f(x)\), which is given as \(x \in [1,5]\). For all values in this domain, we see that \(f(1)=2\) and \(f(5)=10\). Since \(g(y)\) is the inverse of \(f(x)\), we can infer the following information for \(g(y)\): 1. The domain of \(g(y)\) is \(y \in [2,10]\). 2. \(g(2) = 1\) and \(g(10) = 5\).

Calculate the integrals

Now, let's find the value of the expression as given in the problem, which is: $$ \int_{1}^{5} f(x) dx + \int_{2}^{10} g(y) dy $$ We can use the properties of function inverse to simplify this equation. From the relationship mentioned in Step 1, we know that: $$ \int_{1}^{5} f(x) dx + \int_{2}^{10} g(y) dy = \int_{1}^{5} f(x) dx + \int_{f(1)}^{f(5)} g(f(x)) dx $$ Using the substitution rule for integrals, let \(u=f(x)\) and \(du = f'(x)dx\). Then, the equation becomes: $$ \int_{1}^{5} f(x) dx + \int_{f(1)}^{f(5)} g(f(x)) dx = \int_{1}^{5} f(x) dx + \int_{2}^{10} g(u) \left(\frac{du}{f'(x)}\right) $$ Notice that the expression \(\int_{2}^{10} g(u) \left(\frac{du}{f'(x)}\right)\) is equal to \(\int_{1}^{5} x \cdot f'(x)dx\). This is because when \(u=f(x)\) and \(\frac{du}{dx} = f'(x)\), we have: $$ g(u) = g(f(x)) = x $$ Now, our expression becomes: $$ \int_{1}^{5} f(x) dx + \int_{1}^{5} x \cdot f'(x)dx $$ From the fundamental theorem of calculus, the second term in the expression is just the difference between the function values at the limits: $$ \int_{1}^{5} x \cdot f'(x)dx = 5 \cdot f(5) - 1 \cdot f(1) = 5 \cdot 10 - 1 \cdot 2 = 50 - 2 = 48 $$ Adding these integrals, we get: $$ \int_{1}^{5} f(x) dx + \int_{1}^{5} x \cdot f'(x)dx = \int_{1}^{5} f(x) dx + 48 $$ Unfortunately, we cannot find the exact value for \(\int_{1}^{5} f(x) dx\) as the function \(f(x)\) is not given explicitly. However, we have successfully connected the integrals of both functions to each other and simplified it to a single integral plus a constant term.

Key concepts

Inverse Functions

Inverse functions play a vital role in integral calculus and various applications across mathematical studies. In simplest terms, an inverse function reverses the operation of the original function. For a given function f(x), its inverse g(x) will satisfy the conditions g(f(x)) = x and f(g(x)) = x.

Understanding the relationship between a function and its inverse is critical when solving calculus problems like the one presented for IIT JEE aspirants. When the function f(x) and its inverse g(y) are plotted on a graph, their curves are symmetrical with respect to the line y=x. This symmetry can help visualize what's happening when we're working through an integral involving inverse functions.

For example, if we know the values of f(x) at certain points and we're given the range or domain of g(y), we can deduce the values of g(y) at corresponding points due to their interdependent nature. This concept is used in textbook problems to find the relationship between the integrals of a function and its inverse, leading to simplifications and solutions of complex integral problems.

Properties of Integrals

The properties of integrals form the foundation of solving problems in integral calculus. These properties include linearity (integration of sums and scalar multiples), the power rule, the fundamental theorem of calculus, as well as substitution and integration by parts.

A key property often used in combination with inverse functions is the substitution rule, which facilitates the transformation of an integral into a simpler form for evaluation. In our exercise, the substitution \(u=f(x)\) is used to turn the integral of the inverse function into an integral that resembles the original function's integral.

Another property is that integration is antidifferentiation; thus, the integral of a derivative f'(x) corresponding to a continuous function over a closed interval can be solved using the Fundamental Theorem of Calculus. This property states that if you integrate a derivative, you essentially find the change in the values of the original function over that interval. This concept is often utilized to simplify problems that involve inverse functions and their derivatives, as seen in the exercise provided.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is one of the most important equations in all of mathematics, bridging the gap between differentiation and integration. It consists of two parts:
Part 1 states that if a function f is continuous over an interval [a, b] and F is an antiderivative of f over that interval, then \(\int_{a}^{b} f(x) dx = F(b) - F(a)\).
Part 2 asserts that if f is a function that is continuous over an interval and x is any point in that interval, then the function F, defined as \(F(x) = \int_{a}^{x} f(t) dt\), is an antiderivative of f.

In the context of our IIT JEE problem, FTC allows us to evaluate the integral of f'(x) easily without explicitly knowing the form of f(x). By understanding that integrals calculate the accumulated area under the curve, FTC gives us a way to compute this area by finding the net change in the values of function F, which yields a definite value that can be easily interpreted and used in a wider problem-solving context.