Evaluate the integral.

{eq}\displaystyle \int_4^6 \dfrac 1 {\sqrt {t^2 - 9}}\ dt {/eq}

Question:

Evaluate the integral.

{eq}\displaystyle \int_4^6 \dfrac 1 {\sqrt {t^2 - 9}}\ dt {/eq}

Integration:

The area between the curves can be determined by integration. We can simplify the integrals by using trigonometric and inverse trigonometric identities. Finding a function that yields another function on being differentiated is called integration. Using integration, if {eq}\dfrac{d}{dx}\left( F\left( x \right) \right)=f\left( x \right) {/eq}, then {eq}\displaystyle\int{f\left( x \right)dx}=F\left( x \right)+C {/eq}. Some properties of integrals are given below:

  • {eq}\displaystyle\int\limits_{a}^{b}{f\left( x \right)dx}=\displaystyle\int\limits_{a}^{b}{f\left( t \right)dt} {/eq}
  • {eq}\displaystyle\int\limits_{a}^{b}{f\left( x \right)dx}=-\displaystyle\int\limits_{b}^{a}{f\left( x \right)dx} {/eq}
  • {eq}\displaystyle\int\limits_{a}^{b}{f\left( x \right)dx}=\displaystyle\int\limits_{a}^{c}{f\left( x \right)dx}+\displaystyle\int\limits_{c}^{b}{f\left( x \right)dx} {/eq}

Answer and Explanation: 1

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Consider {eq}t=3\sec u {/eq}, so {eq}dt=3\sec u\tan u du {/eq}.

Substitute {eq}4 {/eq} for {eq}t {/eq} into the equation {eq}t=3\sec u {/eq} and...

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Evaluating Definite Integrals Using the Fundamental Theorem

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Chapter 16 / Lesson 2
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In calculus, the fundamental theorem is an essential tool that helps explain the relationship between integration and differentiation. Learn about evaluating definite integrals using the fundamental theorem, and work examples to gain understanding.


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