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Magazine Chapter 6: Problem 28
Evaluate \(\int \sqrt{x-1} x^{2} d x\) by two methods: (a) Make the substitution \(u=x-1 ;(b)\) make the substitution \(v=\sqrt{x}-1 .\)
Short Answer
Expert verified
Using substitution: \(\frac{2}{7}(x-1)^{7/2} + \frac{4}{5}(x-1)^{5/2} + \frac{2}{3}(x-1)^{3/2} + C\) and \(\frac{2}{7}(\sqrt{x})^7 + C\).
Step by step solution
01
Method (a) - Substitution
Choose the substitution \(u = x - 1\). Then \(du = dx\) and \(x = u + 1\). Substitute these into the integral.
02
Rewrite the Integral
The integral \(\int \sqrt{x-1} x^{2} dx\) becomes \(\int \sqrt{u} (u + 1)^{2} du\).
03
Expand and Simplify
Expand \((u + 1)^{2}\) to get \(u^{2} + 2u + 1\). Thus the integral becomes \(\int \sqrt{u} (u^{2} + 2u + 1) du\).
04
Split the Integral
Distribute \(\sqrt{u}\) and split the integral into \(\int u^{5/2} du + 2 \int u^{3/2} du + \int u^{1/2} du\).
05
Integrate
Now integrate each term: \(\int u^{5/2} du = \frac{2}{7}u^{7/2}, 2 \int u^{3/2} du = \frac{4}{5}u^{5/2}, \int u^{1/2} du = \frac{2}{3}u^{3/2}\). Combine the results to get \(\int \sqrt{u} (u^2 + 2u +1) du = \frac{2}{7}u^{7/2} + \frac{4}{5}u^{5/2} + \frac{2}{3}u^{3/2} + C\).
06
Back Substitute
Replace \(u\) back with \(x - 1\) to get \(\frac{2}{7}(x-1)^{7/2} + \frac{4}{5}(x-1)^{5/2} + \frac{2}{3}(x-1)^{3/2} + C\).
07
Method (b) - Substitution
Choose the substitution \(v = \sqrt{x} - 1\). Then \(v = \sqrt{x} - 1\) leads to \(\sqrt{x} = v + 1\), so \(x = (v+1)^{2}\) and \(dx = 2(v+1) dv\).
08
Rewrite the Integral
Substitute \(v, (v+1)^{2},\) and \(2(v+1) dv\) into the integral \(\int \sqrt{x-1} x^{2} dx\). This becomes \(\int v (v+1)^{4} 2(v+1) dv\).
09
Expand and Simplify
Simplify \(2 \int v (v + 1)^5 dv\). Expand to get \(2 \int (v+1)^5 v dv\).
10
Distribute and Integrate
Expand \((v+1)^5\), multiply by \(v\), and integrate term by term. The result is \(\frac{2}{7}(v+1)^{7} + C\).
11
Back Substitute
Replace \(v\) with \(\sqrt{x}-1\). The final result is \(\frac{2}{7}(\sqrt{x})^7 + C\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a fundamental technique in integral calculus. It involves changing the variable of integration to simplify the integral. Here鈥檚 a breakdown of how it works:
- Choose a Substitution: Identify a function within the integral to set as a new variable, such as setting \( u = x - 1 \).
- Find the Differential: Differentiate your substitution to get \( du = dx \).
- Rewrite the Integral: Substitute \( u \) and \( du \) in place of the original variables.
- Integrate and Sub Back: Integrate the simplified integral and then substitute back the original variable.
Definite Integral
A definite integral is evaluated over a specific interval, giving a numerical result:
- Limits of Integration: These are the endpoints of the interval over which you are integrating, typically noted as \( \int_{a}^{b} f(x) dx \).
- Area Under the Curve: The result represents the net area between the function and the x-axis over the interval.
- Fundamental Theorem: The fundamental theorem of calculus connects differentiation and integration, making it possible to evaluate definite integrals by finding antiderivatives.
Indefinite Integral
An indefinite integral, unlike a definite integral, does not have limits of integration. It represents a family of functions:
- General Form: The indefinite integral is written as \( \int f(x) dx \), and it includes a constant of integration noted as \( C \).
- Antiderivative: The result of an indefinite integral is an antiderivative of the original function. Essentially, it is a function whose derivative gives back the original function.
- Constant of Integration: Since differentiation kills constants, the antiderivative must account for all possible constants with \( + C \).
Integration by Parts
Integration by parts is a technique based on the product rule for differentiation. It is used when the integrand is a product of two functions:
- Formula: The integration by parts formula is written as \( \int u dv = uv - \int v du \).
- Choosing \( u \) and \( dv \): A common acronym to help choose which part to differentiate (\( u \)) and which to integrate (\( dv \)) is LIATE (Logarithmic, Inverse Trigonometric, Algebraic, Trigonometric, Exponential).
- Iterative Process: Sometimes, you may need to apply the integration by parts formula more than once to solve an integral.
