Problem 92 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

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) \(100 a\)

Step-by-Step Solution

Answer

B

Step 1: Understanding the Problem

We are given that \( \int_{0}^{100} f(x) \, dx = a \). We need to determine the value of \( \sum_{r=1}^{100}\left(\int_{0}^{1} f(r-1+x) \, dx\right) \). This sum represents integrating the function over 100 different unit intervals.

Step 2: Decompose the Sum

The expression \( \sum_{r=1}^{100}\left(\int_{0}^{1} f(r-1+x) \, dx\right) \) suggests we take the integral of \( f \) from \( r-1 \) to \( r \) for each \( r = 1, 2, \ldots, 100 \) and then sum these integrals. Each integral is over an interval of length 1.

Step 3: Change of Variables

For each term, perform a change of variables. Let \( u = r-1+x \). Then \( x = u-(r-1) \) and \( dx = du \), which means the limits change from \( x = 0 \) to \( x = 1 \) into \( u = r-1 \) to \( u = r \). Therefore, \( \int_{0}^{1} f(r-1+x) \ dx = \int_{r-1}^{r} f(u) \, du \).

Step 4: Recognize the Overall Integral

Summing this from \( r = 1 \) to 100, we have: \( \sum_{r=1}^{100} \int_{r-1}^{r} f(u) \, du = \int_{0}^{1} f(u) \, du + \int_{1}^{2} f(u) \, du + \ldots + \int_{99}^{100} f(u) \, du \), which is equivalent to \( \int_{0}^{100} f(u) \, du \).

Step 5: Substitute Given Value

We already know \( \int_{0}^{100} f(u) \, du = a \) given in the problem. Thus, \( \sum_{r=1}^{100}\left(\int_{0}^{1} f(r-1+x) \, dx\right) = a \).

Key Concepts

Riemann Sum

The concept of the Riemann sum is foundational in understanding how definite integrals work. It acts as an approximation of a definite integral by dividing the area under a curve into thin vertical slices or rectangles. This approximation refines as these slices become thinner, effectively bridging the gap between sums and integrals.

The key steps involve:

• Dividing the interval into a number of smaller subintervals.
• Calculating the function value, or height of each rectangle, at specific points within these subintervals.
• Summing up the areas of all rectangles, which is length times height.

As the number of subintervals increases, or equivalently as the width of these rectangles decreases, the Riemann sum approaches the true value of the definite integral.

Integral Calculus

Integral calculus deals with the concept of finding the accumulation of quantities. While Riemann sums help approximate integrals, integral calculus formalizes this process using precise mathematical definitions.

Some essential elements include:

• The Fundamental Theorem of Calculus: This theorem links differential calculus with integral calculus by illustrating that differentiation and integration are inverse processes.
• Definite integrals: Represent the accumulation of quantities, such as areas under curves, and are expressed with limits of integration.
• Techniques of integration: These are varied methods, such as substitution or integration by parts, employed to solve complex integrals.

In our exercise, recognizing the sum of the integral as equivalent to an overall integral process was crucial to the solution.

Change of Variables

The change of variables, also known as substitution, is a pivotal technique in calculus used to simplify complex integrals. This method involves transforming the original variable into a new one that makes the integration process more straightforward.

To perform a change of variables:

• Select a substitution that simplifies the expression, typically choosing a function inside an integral.
• Determine the differential of the new variable in terms of the original variable.
• Update the integration limits if you're dealing with definite integrals, to reflect the new variable's range.

In the solution provided, the change of variables was used by letting \( u = r-1+x \). This straightforward adjustment allowed the integration over each interval \([r-1, r]\), simplifying the integration process and leading to a clean summation across the main interval \([0, 100]\).