Question

If the function \(f\) is integrable in a closed interval containing \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) and \(\mathrm{d}\), prove that \(\int_{a}^{b} f(x) d x+\int_{b}^{c} f(x) d x+\int_{c}^{d} f(x) d x=\int_{a}^{d} f(x) d x\).

Short answer

Question: Prove that the sum of integrals of the function \(f(x)\) over three sub-intervals \((a,b)\), \((b,c)\), and \((c,d)\) is equal to the integral of \(f(x)\) over the entire interval \((a,d)\). Answer: Using the additive property of definite integrals, we showed that \(\int_{a}^{d} f(x) dx = \int_{a}^{b} f(x) dx + \int_{b}^c f(x) dx + \int_c^d f(x) dx\), proving that the sum of integrals over the three sub-intervals is equal to the integral over the entire interval \((a,d)\).

Step-by-step solution

Write the desired equality using the additive property

According to the additive property, the integral of a function \(f(x)\) over the sum of two non-overlapping intervals is equal to the sum of the integrals of the function on each interval. So the desired equality is: \(\int_{a}^{d} f(x) dx = \int_{a}^{b} f(x) dx + \int_{b}^c f(x) dx + \int_c^d f(x) dx\)

Apply additive property to the left side of the equation

We can rewrite the left side of the equation as the sum of the integrals of \(f(x)\) over two non-overlapping intervals, \((a, c)\) and \((c, d)\), using the additive property as follows: \(\int_{a}^{d} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{d} f(x) dx\)

Apply additive property to the first integral on the right side

We can further apply the additive property to the integral \(\int_{a}^{c} f(x) dx\) by dividing it into two non-overlapping intervals \((a, b)\) and \((b, c)\) as follows: \(\int_{a}^{c} f(x) dx = \int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx\)

Substitute the expression from Step 3 into the equation from Step 2

Now we substitute the expression from Step 3 back into the equation from Step 2: \(\int_{a}^{d} f(x) dx = (\int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx) + \int_{c}^{d} f(x) dx\)

Simplify the equation and conclude the proof

Finally, we simplify the equation and conclude the proof: \(\int_{a}^{d} f(x) dx = \int_{a}^{b} f(x) dx + \int_{b}^c f(x) dx + \int_c^d f(x) dx\) This proves that the sum of integrals of the function \(f\) over three sub-intervals \((a,b)\), \((b,c)\), and \((c,d)\) is equal to the integral of \(f\) over the entire interval \((a,d)\).

Key concepts

Integrable Functions

Understanding the notion of integrable functions is fundamental for solving problems in integral calculus, especially in exams like the IIT JEE. An integrable function is one where it's possible to compute the area under the curve over a given interval. This is crucial for determining definite integrals. In simple terms, if a function can be represented graphically, and the area between the function and the x-axis can be calculated between two points on the graph, that function is considered integrable.

For a function to be integrable in a closed interval, it must be bounded and almost everywhere continuous in that interval. However, there are some functions with points of discontinuity that are still integrable. For IIT JEE aspirants, it's important to be able to identify such functions and apply appropriate integration techniques to solve problems.

Additive Property of Integrals

The additive property of integrals is a critical concept in integral calculus that students must master for exams like the IIT JEE. It states that the integral of a function over a larger interval can be broken down into the sum of integrals over smaller, non-overlapping subintervals. This property is rooted in the notion that the area under a curve from one point to another can be dissected into smaller areas that, when combined, equal the total area.

Using our IIT JEE problem as an example, when we are given three integrals on adjacent intervals and their sum, we can apply the additive property to show that this sum is equal to the integral over the larger interval that encompasses all three.

Visualizing the Additive Property

Think of the function as a path and the integral as the distance traveled on each segment of the path. Splitting the journey into parts and then summing these parts together will give you the total distance, just as the additive property sums the areas over subintervals to find the total area under the curve.

Definite Integrals

Definite integrals play a prominent role in calculus, especially in the context of IIT JEE problems. They are used to calculate the exact area under a curve, between two specified points. This gives them a real-world application in a myriad of fields, from physics to economics, where one needs to find accumulations, areas, or total quantities.

In the context of IIT JEE, understanding definite integrals involves not only knowing the mechanics of integration but also interpreting the integral in terms of areas and accumulations. If we know the definite integral of a function from point 'a' to point 'b', we possess a powerful tool that can give us insights into the behavior of the function over that interval.

To solve problems involving definite integrals for the IIT JEE, it's important that students know the Fundamental Theorem of Calculus, which bridges the concept of antiderivatives with the evaluation of definite integrals, often represented as evaluating the antiderivative at the upper limit of integration and subtracting the value at the lower limit.