Solving the 2020 MIT Integration Bee Qualifier Exam: A Comprehensive Guide

The MIT Integration Bee is a prestigious competition that challenges participants with complex integration problems. In this post, we will explore solutions to several problems from the 2020 MIT Integration Bee Qualifier Exam, demonstrating various integration techniques.

Problem 1: Integral of Logarithmic Functions

Problem Statement

Evaluate the integral:

[ I = \int_{e}^{e^e} \frac{\log x}{x} , dx ]

Solution

To solve this integral, we start by substituting ( u = \log x ). Then, we have:

[ du = \frac{1}{x} , dx \quad \Rightarrow \quad dx = x , du = e^u , du ]

Changing the limits:

• When ( x = e ), ( u = 1 )
• When ( x = e^e ), ( u = e )

Thus, the integral becomes:

[ I = \int_{1}^{e} u , du = \left[ \frac{u^2}{2} \right]_{1}^{e} = \frac{e^2}{2} - \frac{1}{2} = \frac{e^2 - 1}{2} ]

Problem 2: Logarithmic Properties in Integration

Problem Statement

Evaluate the integral:

[ I = \int_{0}^{1} \left( \log(1+x) - \log(1-x) \right) , dx ]

Solution

Using the property of logarithms, we can rewrite the integral as:

[ I = \int_{0}^{1} \log \left( \frac{1+x}{1-x} \right) , dx ]

We can separate this into two integrals:

[ I = \int_{0}^{1} \log(1+x) , dx - \int_{0}^{1} \log(1-x) , dx ]

Using integration by parts for both integrals, we find:

1. For ( \int \log(1+x) , dx ), let ( u = \log(1+x) ) and ( dv = dx ).
2. For ( \int \log(1-x) , dx ), let ( u = \log(1-x) ) and ( dv = dx ).

After evaluating both integrals, we find:

[ I = 2 \log(2) - 1 ]

Problem 3: Integration by Parts with Trigonometric Functions

Problem Statement

Evaluate the integral:

[ I = \int \frac{\sin^{-1} x}{x^3} , dx ]

Solution

We will use integration by parts. Let:

• ( u = \sin^{-1} x )
• ( dv = \frac{1}{x^3} , dx )

Then, we differentiate and integrate:

• ( du = \frac{1}{\sqrt{1-x^2}} , dx )
• ( v = -\frac{1}{2x^2} )

Applying integration by parts:

[ I = -\frac{\sin^{-1} x}{2x^2} + \int \frac{1}{2x^2 \sqrt{1-x^2}} , dx ]

The remaining integral can be solved using a trigonometric substitution. After evaluating, we find:

[ I = -\frac{\sin^{-1} x}{2x^2} + C ]

Problem 4: Integral of Sine and Cosine Functions

Problem Statement

Evaluate the integral:

[ I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \cos(2x) , dx ]

Solution

Using the double angle identity, we can rewrite the integral as:

[ I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin(4x) , dx ]

Evaluating this integral gives:

[ I = -\frac{1}{8} \cos(4x) \bigg|_{0}^{\frac{\pi}{2}} = -\frac{1}{8} (0 - 1) = \frac{1}{8} ]

Conclusion

In this blog post, we explored several integration techniques used to solve problems from the 2020 MIT Integration Bee Qualifier Exam. From logarithmic properties to integration by parts and trigonometric identities, these methods are essential for tackling complex integrals. By practicing these techniques, students can enhance their problem-solving skills and prepare for future mathematical challenges.