Math Problem Statement

Evaluate the integral \( \int_0^5 \frac{dx}{\sqrt{2 - x}} \).

Solution

The image you uploaded contains an example of an integral that needs to be evaluated. The integral expression is:

05dx2x\int_0^5 \frac{dx}{\sqrt{2 - x}}

This is a standard integral of the form dxax\int \frac{dx}{\sqrt{a - x}}, and it can be solved using substitution.

Let's evaluate this integral step by step:

Step 1: Substitution

To simplify the integral, let’s use the substitution:

u=2xu = 2 - x

Then, differentiate both sides with respect to xx:

du=dxdu = -dx

Also, adjust the limits of integration:

  • When x=0x = 0, u=2u = 2.
  • When x=5x = 5, u=3u = -3.

Thus, the integral becomes:

23duu=23u12du\int_2^{-3} \frac{-du}{\sqrt{u}} = - \int_2^{-3} u^{-\frac{1}{2}} du

Step 2: Solving the Integral

Now we integrate u12u^{-\frac{1}{2}}:

23u12du=[2u12]23- \int_2^{-3} u^{-\frac{1}{2}} du = - \left[ 2u^{\frac{1}{2}} \right]_2^{-3}

Step 3: Evaluate the Expression

Substitute the limits:

[2(3)122(2)12]- \left[ 2(-3)^{\frac{1}{2}} - 2(2)^{\frac{1}{2}} \right]

Since 3\sqrt{-3} is undefined in the real number system, this integral likely needs further refinement (perhaps considering complex values). Let me know if you'd like me to guide you through resolving this with complex analysis or a different approach.

Would you like more details on these steps or have any other questions? Here are some related questions for further exploration:

  1. How do you solve integrals with square roots in the denominator using substitution?
  2. What is the concept of complex numbers, and how are they used in integrals like this?
  3. What are common techniques for evaluating definite integrals?
  4. How does the method of substitution help simplify integrals in general?
  5. How do you handle square roots of negative numbers in integrals?

Tip: Always pay attention to the domain of the integrand and adjust your approach accordingly.

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Math Problem Analysis

Mathematical Concepts

Definite Integrals
Substitution Method
Square Root Functions

Formulas

\( \int \frac{dx}{\sqrt{a - x}} = 2\sqrt{a - x} + C \)

Theorems

Substitution in Integration

Suitable Grade Level

College Calculus or Advanced High School