Use integration formulas to get the exact value of \int_{0}^{0.35}\frac{2}{x^2 - 4}dx

Question:

Use integration formulas to get the exact value of

{eq}\int_{0}^{0.35}\frac{2}{x^2 - 4}dx {/eq}

Formula for Integrals:

There are some standard formulas of integration to get the values without substitution and partial fraction method.

The formula for this particular problem is:

  • {eq}\int \frac{1}{x^2-a^2}\ dx= \frac{1}{2a}\ln|\frac{x-a}{x+a}|+C {/eq}

Where,

  • {eq}C {/eq} is the constant of integration and {eq}a {/eq} does not equal to zero.

Answer and Explanation:

The given definite integral is:

{eq}I=\displaystyle \int_{0}^{0.35}\frac{2}{x^2 - 4}\ dx {/eq}

Using the standard fromula of integration, we get:

{eq}\begin{align*} I&=2\displaystyle \int_{0}^{0.35}\frac{1}{x^2 - 2^{2}}\ dx \\ &=2 \left [\frac{1}{2(2)}\ln|\frac{x-2}{x+2}| \right ]_{0}^{0.35}\\ &=2\cdot \frac{1}{4} \left [\ln|\frac{x-2}{x+2}| \right ]_{0}^{0.35}\\ &=\frac{1}{2} \left (\ln|\frac{0.35-2}{0.35+2}|-\ln|\frac{0-2}{0+2}| \right )\\ &=\frac{1}{2} \left (\ln|\frac{-1.65}{2.35}|-\ln|-1| \right )\\ &=\frac{1}{2} \left (\ln\frac{1.65}{2.35}-\ln 1 \right )\\ &=\frac{1}{2} \left (-0.3536-0 \right )\\ &=-0.1768 \end{align*} {/eq}


Learn more about this topic:

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Evaluating Definite Integrals Using the Fundamental Theorem

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2
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