Problem 124 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of the definite integral \(\int_{0}^{1} \frac{x}{x^{2}+16} d x\) lies in the interval \([a, b]\). The smallest such interval is (A) \([0,1]\) (B) \(\left[0, \frac{1}{7}\right]\) (C) \(\left[0, \frac{1}{17}\right]\) (D) None of these

Step-by-Step Solution

Answer

The interval is \(\left[0, \frac{1}{17}\right]\).

Step 1: Understand the problem statement

We need to evaluate the definite integral \(\int_{0}^{1} \frac{x}{x^{2}+16} \, dx\) and identify the interval in which this integral lies. We are given four choices for intervals.

Step 2: Choose a suitable substitution

To solve the integral \(\int_{0}^{1} \frac{x}{x^{2}+16} \, dx\), let's make the substitution \(u = x^2 + 16\), which implies \(du = 2x \, dx\). Thus, \(x \, dx = \frac{du}{2}\).

Step 3: Change the limits of integration

Using the substitution \(u = x^2 + 16\), we change the limits of integration. When \(x = 0\), \(u = 16\), and when \(x = 1\), \(u = 17\). The new integral becomes \(\int_{16}^{17} \frac{1}{u} \cdot \frac{du}{2}\).

Step 4: Simplify and compute the integral

The integral \(\int_{16}^{17} \frac{1}{u} \cdot \frac{du}{2}\) simplifies to \(\frac{1}{2} \int_{16}^{17} \frac{1}{u} \, du\). This is a standard integral and its solution is \(\frac{1}{2} \ln|u| \), evaluated from 16 to 17.

Step 5: Evaluate the integral at the new limits

Evaluate \(\frac{1}{2} \ln|u|\) from 16 to 17: \[ \frac{1}{2} (\ln|17| - \ln|16|) = \frac{1}{2} \ln \left(\frac{17}{16}\right) \].

Step 6: Approximate and compare with intervals

Calculate \(\frac{1}{2} \ln \left(\frac{17}{16}\right) \). This value is approximately 0.0301. Compare this with the intervals provided: it lies within the interval \(\left[0, \frac{1}{17}\right]\).

Step 7: Conclusion

The smallest interval that contains the value of the definite integral is \(\left[0, \frac{1}{17}\right]\).

Key Concepts

Substitution Method

The substitution method is a technique for simplifying the integration process, especially helpful for definite integrals. Here, we substitute one part of the integrand (the function under the integral sign) with a new variable.

Think of it like changing the path to a problem to an easier one. In our problem, we had the integral \(\int_{0}^{1} \frac{x}{x^{2}+16} \, dx\). This form can be tricky to handle directly.

To make it easier, we substitute \(u = x^2 + 16\), which turns our function into a more manageable form. With this substitution, we also need the derivative: \(du = 2x \, dx\). This allows us to replace \(x \, dx\) with \(\frac{du}{2}\).

There are some steps to always remember when using the substitution method:

• Identify the part of the integrand to substitute.
• Express \(dx\) in terms of \(du\).
• Change the limits of integration accordingly.

Implementing this strategy results in converting the original integral into a form that is often easier to integrate.

Definite Integral Evaluation

Once the substitution is made, the next step is to evaluate the definite integral, which now has different limits and a transformed integrand.

For our integral, the substitution transformed it into \(\frac{1}{2} \int_{16}^{17} \frac{1}{u} \, du\). At this point, the integration is simplified to a known formula. The integral of \(\frac{1}{u}\) is \(\ln|u|\). Thus, we process the definite integral as:\[ \frac{1}{2} ( \ln|u| \bigg|_{16}^{17} ) = \frac{1}{2} ( \ln|17| - \ln|16| ) = \frac{1}{2} \ln \left( \frac{17}{16} \right). \]
The evaluation process involves substituting the upper limit and subtracting the value at the lower limit. Applying these steps allows us to find precise values even when dealing with complex integrals.

Remember that the goal is to find the net area under the curve between two points on a graph, which is what definite integrals represent.

Limits of Integration

Changing the limits of integration is a crucial aspect of substitution. It ensures that the transformed integral corresponds to the same area under the curve as the original.

In our exercise, the original limits for \(x\) were from 0 to 1. After substituting \(u = x^2 + 16\), we found:

• When \(x = 0\), \(u = 16\).
• When \(x = 1\), \(u = 17\).

These new limits \(16\) to \(17\) are used in the integral \(\frac{1}{2} \int_{16}^{17} \frac{1}{u} \, du\). It's important to replace the original limits with the new ones correctly to maintain the integrity of the integral's value.

By doing this, we ensure our substitutions and simplifications are valid throughout the calculation process. Always double-check that the limits align with your substitution for a successful integration.