๐ U-Substitution with Definite Integrals: A Comprehensive Guide
U-substitution is a powerful technique for evaluating integrals, especially when dealing with composite functions. When applying it to definite integrals, a crucial step is correctly handling the limits of integration. There are two main approaches: either transform the integral back to the original variable and use the original limits, or change the limits to correspond to the new variable. We'll focus on the latter, which is often more efficient.
๐ History and Background
The method of u-substitution is rooted in the chain rule for differentiation. The concept gained prominence with the development of calculus in the 17th century by Newton and Leibniz. Over time, mathematicians formalized the technique for both indefinite and definite integrals. U-substitution simplifies complex integrals, making them solvable within reasonable timeframes.
๐ Key Principles
- ๐ Choose a Suitable 'u': Identify a part of the integrand (the function being integrated) that, when substituted with 'u', simplifies the integral. Often, this is the inner function of a composite function.
- ๐ Compute du: Find the derivative of 'u' with respect to 'x' (i.e., $\frac{du}{dx}$) and solve for $dx$ in terms of $du$.
- ๐ Substitute: Replace the chosen part of the integrand with 'u' and $dx$ with its equivalent expression in terms of $du$.
- ๐ Change the Limits of Integration: This is the critical step! If the original integral has limits $a$ and $b$ (i.e., $\int_{a}^{b}$), you need to find the corresponding limits for 'u'. Calculate $u(a)$ and $u(b)$. These become the new limits.
- โ Evaluate the New Integral: Evaluate the transformed integral with the new limits: $\int_{u(a)}^{u(b)} f(u) du$.
โ๏ธ Step-by-Step Process with Example
Let's consider the definite integral: $\int_{0}^{2} x \sqrt{x^2 + 1} dx$.
- Choose u: Let $u = x^2 + 1$.
- Compute du: $\frac{du}{dx} = 2x$, so $dx = \frac{du}{2x}$.
- Substitute: The integral becomes $\int x \sqrt{u} \frac{du}{2x} = \frac{1}{2} \int \sqrt{u} du$.
- Change the Limits:
- When $x = 0$, $u = 0^2 + 1 = 1$.
- When $x = 2$, $u = 2^2 + 1 = 5$.
- Evaluate: The new integral is $\frac{1}{2} \int_{1}^{5} u^{1/2} du = \frac{1}{2} [\frac{2}{3}u^{3/2}]_{1}^{5} = \frac{1}{3} [u^{3/2}]_{1}^{5} = \frac{1}{3} (5^{3/2} - 1^{3/2}) = \frac{1}{3} (5\sqrt{5} - 1)$.
๐ก Common Errors to Avoid
- ๐ซ Forgetting to Change the Limits: This is the most common mistake. Remember to find the new limits in terms of 'u'.
- โ Incorrectly Calculating 'du': Double-check your differentiation when finding $\frac{du}{dx}$.
- ๐ Not Simplifying After Substitution: Simplify the integral after substitution to make it easier to evaluate.
- ๐คฏ Using Original Limits After Substitution: Never use the x-limits on a u-integral. Always convert, or convert back after integrating.
โ๏ธ Real-world Examples
- ๐ Physics: Calculating the work done by a variable force, where the force is a function of displacement.
- ๐ Statistics: Finding probabilities associated with certain probability density functions.
- ๐ฐ Economics: Determining consumer surplus or producer surplus.
๐ค Practice Quiz
| Question |
Answer |
| $\int_{0}^{1} 2x(x^2 + 1)^3 dx$ |
$\frac{15}{2}$ |
| $\int_{0}^{\pi/2} \sin(x) \cos^2(x) dx$ |
$\frac{1}{3}$ |
| $\int_{1}^{e} \frac{\ln(x)}{x} dx$ |
$\frac{1}{2}$ |
| $\int_{0}^{\sqrt{\pi}} x \sin(x^2) dx$ |
$1$ |
| $\int_{0}^{1} x e^{x^2} dx$ |
$\frac{1}{2}(e-1)$ |
| $\int_{0}^{\pi/4} \tan(x) \sec^2(x) dx$ |
$\frac{1}{2}$ |
| $\int_{1}^{2} \frac{x}{x^2+1} dx$ |
$\frac{1}{2} \ln(\frac{5}{2})$ |
โ
Conclusion
Mastering u-substitution with definite integrals and new bounds requires understanding the underlying principles and avoiding common errors. By carefully changing the limits of integration, you can efficiently evaluate these integrals and apply them to various real-world scenarios. Keep practicing, and you'll become proficient in no time!