Problem 37 - Calculus Early Transcendentals

Question

Evaluate the following integrals or state that they diverge. $$\int_{1}^{2} \frac{d x}{\sqrt{x-1}}$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral $$\int_{1}^{2} \frac{d x}{\sqrt{x-1}}$$. Answer: 2

Step 1: Identify the substitution

We notice that the integral involves a square root in the denominator and the expression inside the square root is linear (\(x-1\)). We can perform a substitution to simplify the integral. Let's let \(u = x-1\). Then, \(d u = d x\) and the limits of integration will change as well.

Step 2: Change limits of integration

When \(x = 1\), \(u = 1 - 1 = 0\). When \(x = 2\), \(u = 2 - 1 = 1\). So the new limits of integration are from \(0\) to \(1\).

Step 3: Rewrite the integral using the substitution

Now that we have our substitution \(u = x-1\) and the new limits of integration, we can rewrite the integral as: $$\int_{0}^{1} \frac{d u}{\sqrt{u}}$$

Step 4: Evaluate the integral

We can rewrite the integral in terms of powers of \(u\): $$\int_{0}^{1} u^{-\frac{1}{2}} d u$$ Now let's find the antiderivative. To do this, we'll add \(1\) to the power and divide by the new power: $$\frac{u^{\frac{1}{2}}}{\frac{1}{2}}$$ Let's evaluate this antiderivative at the limits of integration: $$\left[\frac{u^{\frac{1}{2}}}{\frac{1}{2}}\right]_0^1$$ Now substitute the limits of integration: $$\frac{1^{\frac{1}{2}}}{\frac{1}{2}} - \frac{0^{\frac{1}{2}}}{\frac{1}{2}}$$ Finally, simplify the expression: $$2(1)-2(0) = 2$$ The final result of the definite integral is: $$\int_{1}^{2} \frac{d x}{\sqrt{x-1}}=2$$.