Question:medium

\[ \int_0^1 x\sqrt{x^2+4}\,dx= \]

Show Hint

For integrals containing \(x\sqrt{x^2+a}\), use \(u=x^2+a\).
  • \(\frac{1}{3}[5\sqrt{5}-4]\)
  • \(\frac{1}{2}[5\sqrt{5}-8]\)
  • \(\frac{1}{3}[5\sqrt{5}-8]\)
  • \(\frac{1}{3}[5\sqrt{5}+4]\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
This is a definite integral that can be solved using the substitution method.
The integrand is a product of \( x \) and a function of \( x^2 \), which is a perfect scenario for substitution because the derivative of \( x^2 \) is proportional to \( x \).
Step 2: Key Formula or Approach:
1. Use substitution \( t = x^2 + 4 \).
2. Find the new limits of integration.
3. Apply the power rule.
Step 3: Detailed Explanation:

Step 3.1: Substitution:
Let \( t = x^2 + 4 \).
Differentiate: \( dt = 2x dx \implies x dx = \frac{dt}{2} \).

Step 3.2: Change of Limits:
When \( x = 0 \), \( t = 0^2 + 4 = 4 \).
When \( x = 1 \), \( t = 1^2 + 4 = 5 \).

Step 3.3: Integral Transformation:
The integral \( \int_0^1 x \sqrt{x^2 + 4} dx \) becomes:
\[ \int_4^5 \sqrt{t} \frac{dt}{2} = \frac{1}{2} \int_4^5 t^{1/2} dt \]

Step 3.4: Evaluation:
\[ \frac{1}{2} \left[ \frac{t^{3/2}}{3/2} \right]_4^5 = \frac{1}{2} \cdot \frac{2}{3} \left[ t^{3/2} \right]_4^5 \]
\[ = \frac{1}{3} [ 5^{3/2} - 4^{3/2} ] \]

Step 3.5: Final Simplification:
\( 5^{3/2} = 5 \sqrt{5} \).
\( 4^{3/2} = ( \sqrt{4} )^3 = 2^3 = 8 \).
Result: \( \frac{1}{3} [5\sqrt{5} - 8] \).

Step 4: Final Answer:
The result of the definite integral is \( \frac{1}{3} [5\sqrt{5} - 8] \).
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