Math Problem Statement

Evaluate the definite integral \( \int_{0}^{\sqrt{3}} \frac{4x}{\sqrt{x^2 + 1}} dx \).

Solution

The given problem is to evaluate the definite integral:

034xx2+1dx\int_{0}^{\sqrt{3}} \frac{4x}{\sqrt{x^2 + 1}} \, dx

Step-by-step Solution:

  1. Substitute: Let u=x2+1u = x^2 + 1, so that du=2xdxdu = 2x \, dx.

    • When x=0x = 0, u=02+1=1u = 0^2 + 1 = 1.
    • When x=3x = \sqrt{3}, u=(3)2+1=4u = (\sqrt{3})^2 + 1 = 4.

    Rewrite the integral in terms of uu:

    034xx2+1dx=144udu2=142udu\int_{0}^{\sqrt{3}} \frac{4x}{\sqrt{x^2 + 1}} \, dx = \int_{1}^{4} \frac{4}{\sqrt{u}} \cdot \frac{du}{2} = \int_{1}^{4} \frac{2}{\sqrt{u}} \, du

  2. Simplify: The integral simplifies to:

    142udu=214u12du\int_{1}^{4} \frac{2}{\sqrt{u}} \, du = 2 \int_{1}^{4} u^{-\frac{1}{2}} \, du

  3. Integrate: Using the power rule undu=un+1n+1+C\int u^n \, du = \frac{u^{n+1}}{n+1} + C, we have:

    2u12du=2[2u12]142 \int u^{-\frac{1}{2}} \, du = 2 \cdot \left[ 2u^{\frac{1}{2}} \right]_{1}^{4}

    =4[u12]14= 4 \left[ u^{\frac{1}{2}} \right]_{1}^{4}

  4. Evaluate: Substitute the limits:

    [ 4 \left[ u^{\frac{1}{2}} \right]_{1}^{4} = 4 \left[ \sqrt{4} - \sqrt{1} \right] = 4 \left[ 2 - 1 \right] = 4 \cdot 1 = 4 ]

Final Answer:

4\boxed{4}


Do you want a detailed explanation of any step?

Related Questions:

  1. What is substitution in integration, and why is it useful?
  2. How does the power rule work for definite integrals?
  3. What is the geometric interpretation of this integral?
  4. How would the limits change if xx had a different range?
  5. What happens if the integral bounds are reversed (e.g., from 3\sqrt{3} to 00)?

Tip:

When integrating rational functions with square roots, substitution often simplifies the integral significantly!

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Math Problem Analysis

Mathematical Concepts

Definite Integrals
Substitution Method in Integration
Simplification of Rational Functions with Square Roots

Formulas

Substitution: Let \( u = g(x) \), \( du = g'(x) dx \).
Definite Integral with substitution: \( \int_a^b f(x) dx = \int_{g(a)}^{g(b)} f(u) \cdot \frac{du}{g'(x)} \)
Power Rule for Integration: \( \int u^n du = \frac{u^{n+1}}{n+1} + C \) for \( n \neq -1 \)

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12 or early college-level calculus