Problem 12 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int_{-5}^{0} \frac{d x}{\sqrt{4-x}}$$

Step-by-Step Solution

Answer

Question: Evaluate the integral: \(\int_{-5}^{0} \frac{d x}{\sqrt{4-x}}\) Answer: The value of the definite integral is 2.

Step 1: Select a substitution

Let's make a substitution which simplifies the given expression. Let \(u = 4 - x\). Now we need to find the derivative of \(u\) with respect to \(x\). \(\frac{du}{dx} = \frac{d}{dx}(4 - x)=-1\)

Step 2: Replace \(dx\)

We need to express \(dx\) in terms of \(du\). To do that, we have: \(dx = -du\) Now let's substitute \(u\) and \(-du\) for \(4-x\) and \(dx\) respectively: $$\int_{-5}^{0} \frac{-du}{\sqrt{u}}$$

Step 3: Change integration limits

To make the substitution complete, we need to change the integration limits as well. Since \(u = 4 - x\), for \(x = -5\), the lower limit, we have \(u = 4 - (-5) = 9\) and for \(x = 0\), the upper limit we have \(u = 4 - 0 = 4\) The integral now becomes: $$\int_{9}^{4} \frac{-du}{\sqrt{u}}$$

Step 4: Evaluate the integral

This integral is simpler to compute, as it is a basic power function. We can rewrite the integral as: $$-\int_{9}^{4} u^{-\frac{1}{2}}du$$ Now, we can use the power rule for integration: $$-\left[u^{\frac{1}{2}}\left(\frac{1}{\frac{1}{2}}\right)\right]_9^4 = -2\left(u^{\frac{1}{2}}\right)_9^4$$ Applying the limits of integration, $$-2\left(\sqrt{4} - \sqrt{9}\right)$$

Step 5: Simplify the result

Finally, we can simplify the expression to get the final result: $$-2(2-3) = -2(-1) = 2$$ So, the value of the definite integral is 2: $$\int_{-5}^{0} \frac{d x}{\sqrt{4-x}}=2$$