Evaluate the definite integral ∫_{-1}^{1} (5r / (4+r²)²) dr
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Let's evaluate the definite integral of 5r over (4 plus r squared) squared, from negative 1 to 1. First, we'll examine the integrand to see if we can apply any special techniques.
To solve this integral efficiently, let's check if the integrand is an odd function. A function is odd if f of negative r equals negative f of r. Substituting negative r into our function, we get 5 times negative r over 4 plus negative r squared, all squared. Since negative r squared is just r squared, this simplifies to negative 5r over 4 plus r squared squared, which is exactly negative f of r. Therefore, our function is odd.
Since we've established that our integrand is an odd function, we can use a special property of definite integrals. When an odd function is integrated over symmetric limits around zero, the integral equals zero. This is because the area under the curve on the positive side exactly cancels out the area on the negative side. In our case, with limits from negative one to one, this property directly applies. Therefore, the value of our integral is zero.
To verify our result, let's solve the integral using the substitution method. We'll let u equal 4 plus r squared. Taking the derivative, we get du equals 2r dr, which means r dr equals one-half du. Substituting into our integral, we have the integral of 5 over u squared times one-half du. This simplifies to five-halves times the integral of u to the negative two du. Integrating, we get five-halves times negative u to the negative one, which is negative 5 over 2u. Substituting back for u, we obtain negative 5 over 2 times 4 plus r squared. Now we evaluate at our bounds: F of 1 equals negative 1 half, and F of negative 1 also equals negative 1 half. Therefore, F of 1 minus F of negative 1 equals zero, confirming our earlier result.
In conclusion, we've evaluated the definite integral of 5r over (4 plus r squared) squared from negative 1 to 1. Using the property of odd functions, we immediately determined that the integral equals zero. We then verified this result using the substitution method, which also yielded zero. Both approaches confirm that when integrating an odd function over symmetric limits, the positive and negative areas exactly cancel each other out, resulting in a total value of zero.