The integral in this exercise converges. Evaluate the integral without using a table. $$ \int_{0}^{9} \frac{d x}{\sqrt{9-x}} $$

Question

UpStudy AI · Accepted Answer

To evaluate the integral

$$
\int_{0}^{9} \frac{dx}{\sqrt{9 - x}},
$$

we can use a substitution method.

**Step 1: Substitution**

Let’s set:

$$
u = 9 - x.
$$

Then, the differential $$du$$ is:

$$
du = -dx \quad \Rightarrow \quad dx = -du.
$$

**Step 2: Change of Limits**

When $$x = 0$$:
$$
u = 9 - 0 = 9.
$$

When $$x = 9$$:
$$
u = 9 - 9 = 0.
$$

**Step 3: Rewrite the Integral**

Substituting $$u$$ and $$dx$$ into the integral, we get:

$$
\int_{0}^{9} \frac{dx}{\sqrt{9 - x}} = \int_{9}^{0} \frac{-du}{\sqrt{u}} = \int_{0}^{9} \frac{du}{\sqrt{u}}.
$$

**Step 4: Evaluate the Integral**

The integral simplifies to:

$$
\int_{0}^{9} u^{-1/2} \, du.
$$

The antiderivative of $$u^{-1/2}$$ is:

$$
2u^{1/2}.
$$

Evaluating from 0 to 9:

$$
2\sqrt{u} \bigg|_{0}^{9} = 2\sqrt{9} - 2\sqrt{0} = 2 \times 3 - 0 = 6.
$$

**Final Answer:**

6
The integral evaluates to 6.

The integral in this exercise converges. Evaluate the integral without using a table. \( \int_{0}^{9} \frac{d x}{\sqrt{9-x}} \)