Question

Express the integral \(\int_{2}^{6} \frac{\mathrm{x}}{1+\mathrm{x}^{5}} \mathrm{~d} \mathrm{x}\) as a limit of integral sums. Do not evaluate the limit.

Short answer

Question: Express the integral \(\int_{2}^{6} \frac{x}{1+x^{5}} dx\) as a limit of integral sums. Answer: The integral \(\int_{2}^{6} \frac{x}{1+x^{5}} dx\) can be expressed as the limit of integral sums as follows: $$\int_{2}^{6} \frac{x}{1+x^{5}} dx = \lim_{n\to\infty} \sum_{i=1}^{n} \frac{2 + (i-\frac{1}{2})\frac{4}{n}}{1 + (2 + (i-\frac{1}{2})\frac{4}{n})^{5}} \cdot \frac{4}{n}$$

Step-by-step solution

Partition the interval [2, 6]

Divide the interval [2, 6] into n equal subintervals. The width of each subinterval is \(\Delta x = \frac{6 - 2}{n} = \frac{4}{n}\).

Choose a representative value x_i in each subinterval

Let \(x_i\) represent a value in the \(i\)-th subinterval [\(2+(i-1)\Delta x\), \(2+i\Delta x\)]. We can choose \(x_{i} = 2 + (i-\frac{1}{2})\Delta x\).

Express the Riemann sum

The Riemann sum for the integral \(\int_{2}^{6} \frac{x}{1+x^{5}} dx\) is given by: $$\sum_{i=1}^{n} f(x_{i}) \Delta x$$ Substitute \(f(x) = \frac{x}{1+x^{5}}\) and the chosen \(x_i\), and find the Riemann sum. $$\sum_{i=1}^{n} \frac{2 + (i-\frac{1}{2})\Delta x}{1 + (2 + (i-\frac{1}{2})\Delta x)^{5}} \cdot \Delta x$$

Express the integral as a limit of integral sums

Now, we rewrite the integral as the limit of the Riemann sum as n approaches infinity: $$\int_{2}^{6} \frac{x}{1+x^{5}} dx = \lim_{n\to\infty} \sum_{i=1}^{n} \frac{2 + (i-\frac{1}{2})\Delta x}{1 + (2 + (i-\frac{1}{2})\Delta x)^{5}} \cdot \Delta x$$ The desired expression for the integral is the limit of the integral sums: $$\int_{2}^{6} \frac{x}{1+x^{5}} dx = \lim_{n\to\infty} \sum_{i=1}^{n} \frac{2 + (i-\frac{1}{2})\frac{4}{n}}{1 + (2 + (i-\frac{1}{2})\frac{4}{n})^{5}} \cdot \frac{4}{n}$$

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with the accumulation of quantities and the areas under and between curves. When we talk about computing an integral, we're often referring to finding the area under a function's graph on a given interval. One powerful technique of integral calculus is integration by Riemann sums.

A Riemann sum is a method of approximating the integral of a function by dividing the area under the curve into small rectangles and summing their areas. As the number of rectangles increases, the approximation improves. Ultimately, the exact area under the curve is obtained in the limit as the number of rectangles approaches infinity.

The step-by-step solution provided for the given exercise is an example of a Riemann sum. The exercise involves partitioning the interval and selecting points within each partition to form rectangles. By summing the products of each rectangle's height, given by the function \(f(x_i)\), and its width \(\Delta x\), the approximate area under the function on the interval from 2 to 6 is calculated. This sum is then expressed as a limit to find the exact value of the integral.

Limits

In calculus, the concept of a limit is fundamental. It captures the idea of approximation and the behavior of functions as inputs approach certain values. Limits help us understand the behavior of functions at points where they might not be well-defined by considering the values that the function approaches as the variable gets closer to a particular point.

When we discussed expressing the integral as a limit of Riemann sums in the previous section, we were leading into the concept of limits. The phrase \(\lim_{n\to\infty}\) symbolizes the action of taking the limit as the number of subintervals \(n\) becomes very large. In our exercise, the limit represents the exact value of the definite integral as the width of our rectangles \(\Delta x\) trends towards zero, refining the approximation to perfection. Limits are essential in the practice of integral calculus and are at the heart of defining integrals themselves.

IIT JEE Main and Advanced Mathematics

The Indian Institutes of Technology's Joint Entrance Examination (IIT JEE) Main and Advanced are highly competitive exams aspirants take to secure a seat in one of India’s prestigious engineering colleges. These exams test the depth of understanding and application of several core concepts in mathematics, including integral calculus and limits.

Students preparing for IIT JEE must be proficient in a wide range of mathematical techniques, importantly Riemann sums and the concept of limits, as demonstrated in the exercise's step-by-step solution. IIT JEE problems often involve interpretive and application-based questions where a deep understanding of the underlying principles of calculus is essential. The ability to express integrals as the limit of sums and to manipulate these expressions is a vital skill in achieving success on the engineering entrance exams.