Math Problem Statement

integral from 0 to 4 of (2x)/(sqrt(xx + 9))dx using the substitution u = sqrt(xx + 9)

Solution

To solve the integral 042xx2+9dx\int_0^4 \frac{2x}{\sqrt{x^2 + 9}} \, dx using the substitution u=x2+9u = \sqrt{x^2 + 9}, follow these steps:

Step 1: Substitution

Given the substitution u=x2+9u = \sqrt{x^2 + 9}, we first need to express dudu in terms of dxdx.

u=x2+9u = \sqrt{x^2 + 9} Square both sides: u2=x2+9u^2 = x^2 + 9 Differentiate both sides with respect to xx: 2ududx=2x2u \frac{du}{dx} = 2x This simplifies to: dudx=xu\frac{du}{dx} = \frac{x}{u} Therefore: du=xudxdu = \frac{x}{u} dx Substitute u=x2+9u = \sqrt{x^2 + 9} back: du=xx2+9dxdu = \frac{x}{\sqrt{x^2 + 9}} dx Now solve for dxdx: dx=uduxdx = \frac{u \, du}{x} But since x=u29x = \sqrt{u^2 - 9} (from u2=x2+9u^2 = x^2 + 9), we can write: dx=uduu29dx = \frac{u \, du}{\sqrt{u^2 - 9}}

Step 2: Rewrite the Integral

Now, substitute into the integral:

042xx2+9dx=u(0)u(4)2u29uuduu29\int_0^4 \frac{2x}{\sqrt{x^2 + 9}} \, dx = \int_{u(0)}^{u(4)} \frac{2\sqrt{u^2 - 9}}{u} \cdot \frac{u \, du}{\sqrt{u^2 - 9}}

Simplify the expression: u(0)u(4)2du\int_{u(0)}^{u(4)} 2 \, du

Step 3: Update Limits of Integration

We need to change the limits of integration according to the substitution u=x2+9u = \sqrt{x^2 + 9}:

  • When x=0x = 0: u(0)=02+9=9=3u(0) = \sqrt{0^2 + 9} = \sqrt{9} = 3.
  • When x=4x = 4: u(4)=42+9=16+9=25=5u(4) = \sqrt{4^2 + 9} = \sqrt{16 + 9} = \sqrt{25} = 5.

So the integral becomes: 352du\int_3^5 2 \, du

Step 4: Integrate

Now, integrate with respect to uu:

352du=2u35=2(5)2(3)=106=4\int_3^5 2 \, du = 2u \Big|_3^5 = 2(5) - 2(3) = 10 - 6 = 4

Final Answer

The value of the integral is 4\boxed{4}.


Would you like more details on any part of this solution? Here are some related questions that might interest you:

  1. How do you handle substitution when the integral bounds change?
  2. What are common substitutions for integrals involving square roots?
  3. How does the choice of substitution affect the complexity of solving an integral?
  4. What is the geometric interpretation of the integral 042xx2+9dx\int_0^4 \frac{2x}{\sqrt{x^2 + 9}} dx?
  5. How can we verify the correctness of an integral using differentiation?

Tip: Always double-check the limits of integration after performing a substitution, as they change with the new variable.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration
Substitution Method

Formulas

Substitution Method: u = sqrt(x^2 + 9)
Integral of a function using substitution

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12