Problem 23 - 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) a

Step 1: Understand the problem

We are given that \(\int_{0}^{100} f(x)\, dx = a\) and need to evaluate the sum \(\sum_{r=1}^{100}\left(\int_{0}^{1} f(r-1+x)\, dx\right)\). We will simplify the expression and relate it to the given integral.

Step 2: Rewrite the sum of integrals

Each term in the sum \(\sum_{r=1}^{100}\left(\int_{0}^{1} f(r-1+x)\, dx\right)\) can be rewritten using a variable substitution. Consider \( y = r-1+x\), which means \( dy = dx\) and when \( x = 0\), \( y = r - 1\) and when \( x = 1\), \( y = r\). So, \(\int_{0}^{1} f(r-1+x)\, dx = \int_{r-1}^{r} f(y)\, dy\).

Step 3: Sum the rewritten integrals

The expression \(\sum_{r=1}^{100}\left(\int_{r-1}^{r} f(y)\, dy\right)\) represents the integral of \(f(y)\) over the interval from \(0\) to \(100\). Conditions ensure these intervals are adjacent and complete. Therefore, \(\int_{r-1}^{r} f(y)\, dy\) over all \(r\) from \(1\) to \(100\) is same as \(\int_{0}^{100} f(y)\, dy\).

Step 4: Final evaluation of the expression

Since \(\int_{0}^{100} f(y)\, dy = a\) is given, the sum \(\sum_{r=1}^{100}\left(\int_{r-1}^{r} f(y)\, dy\right) = \int_{0}^{100} f(y)\, dy = a\). So, \(\sum_{r=1}^{100}\left(\int_{0}^{1} f(r-1+x)\, dx\right) = a\).

Key Concepts

Integral Calculus

Integral Calculus is a branch of mathematics focused on integrals and their applications. It deals with concepts like finding areas under curves and accumulation of quantities. When we talk about a definite integral, such as \( \int_{a}^{b} f(x)\,dx \), it represents the area under the curve of the function \( f(x) \) from \( x = a \) to \( x = b \). This concept is crucial for solving real-world problems involving rates of change and accumulation over intervals.

Integral Calculus helps unify and generalize calculations of areas and volumes. It essentially reverses what is done in differential calculus. While differentiation breaks down rates of change, integration combines rates to determine total value.
Some properties of definite integrals include:

• Additivity: \( \int_{a}^{b} f(x)\, dx + \int_{b}^{c} f(x)\, dx = \int_{a}^{c} f(x)\, dx \).
• Linearity: \( \int_{a}^{b} [cf(x) + g(x)]\, dx = c \int_{a}^{b} f(x)\, dx + \int_{a}^{b} g(x)\, dx \).

Understanding these properties helps simplify complex integrals and applies to scenarios like the exercise above, where we break complex summations into simpler forms.

Variable Substitution

Variable substitution, also known as change of variables, is a method used to simplify integrals. This technique involves changing the variable of integration to transform a difficult integral into a simpler one. It is akin to using the chain rule in differentiation but in reverse.

For example, in the exercise, each integral in the summation \( \sum_{r=1}^{100} (\int_{0}^{1} f(r-1+x)\, dx) \) is transformed using the substitution \( y = r-1+x \), \( dy = dx \). This changes the limits of integration accordingly: when \( x = 0 \), \( y = r-1 \), and when \( x = 1 \), \( y = r \). Thus, the integral \( \int_{0}^{1} f(r-1+x)\, dx \) becomes \( \int_{r-1}^{r} f(y)\, dy \).

This method helps across many integral problems, particularly when dealing with composite functions or integrands that involve shifting patterns or repeating intervals. It simplifies the problem, allowing us to focus on the transformed, often simpler, integral or function.

Summation of Integrals

The summation of integrals is a powerful concept in calculus, allowing for the algebraic sum of several integrals to form contiguous or composite integral forms. In problems involving a sum of integrals like \( \sum_{r=1}^{100} \left( \int_{r-1}^{r} f(y)\, dy \right) \), we observe the collective accumulation of each individual integral over the specified intervals.

When tackled methodically, such a sum is essentially merging smaller integrals into a larger, single integral provided there are no gaps or overlaps in the integration intervals. In the exercise, \( \sum_{r=1}^{100} \left( \int_{r-1}^{r} f(y)\, dy \right) \) covers the entire span from \( 0 \) to \( 100 \), thus it equals \( \int_{0}^{100} f(y)\, dy \).

This technique is useful when analyzing periodic or piecewise functions. Understanding how to apply this allows for efficient computation of net effects over ongoing processes, captured plainly in the integral framework. With known initial intervals, the entire area under the curve can be computed succinctly.