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Evaluate the integral \int_1^3 \frac{2}{\sqrt{15 + 2x - x^2}} \,dx

Question:

Evaluate the integral

{eq}\int_1^3 \frac{2}{\sqrt{15 + 2x - x^2}} \,dx {/eq}

Integral:

To evaluate the given integral, we complete the square in the denominator.

This puts the given definite integral into the form of {eq}\int f(g(x))g'(x)dx {/eq}. Then, we'll use the substitution method to solve the integral. We'll differentiate the obtained substitution. As we are dealing with definite integral, we also need to update the limits and back substitute the values in the given expression and simplify it.

Answer and Explanation:

Given:

{eq}\int_1^3 \dfrac{2}{\sqrt{15 + 2x - x^2}} \,dx {/eq}

We have to solve the integral.


{eq}\begin{align} \int_1^3 \dfrac{2}{\sqrt{15 + 2x - x^2}} \,dx &=\int_1^3 \dfrac{2}{\sqrt{15 + 2x - x^2+1-1}} \,dx\\ &=\int_1^3 \dfrac{2}{\sqrt{16 -(x^2-2x+1)}} \,dx\\ &=\int_1^3 \dfrac{2}{\sqrt{16 - (x-1)^2}} \,dx \end{align} {/eq}


We use the substitution {eq}x-1=t {/eq} to solve the integral.


Differentiate both sides with respect to {eq}x {/eq}

{eq}\begin{align} 1-0 &=\dfrac{dt}{dx}\\ dx &=dt \end{align} {/eq}


As we are dealing with the definite integral, we also need to update the limits of integration according to the substitution.

When:

{eq}x=1 \Rightarrow t=1-1=0\\ x=3 \Rightarrow t=3-1=2 {/eq}


Applying these to the integral, we have:

{eq}\begin{align} \int_1^3 \dfrac{2}{\sqrt{16 - (x-1)^2}} \,dx &=\int_0^2 \dfrac{2}{\sqrt{16 - t^2}} \,dt\\ &=2\int_0^2 \dfrac{1}{\sqrt{16 - t^2}} \,dt\ & \left [ \because \int af(x)dx=a\int f(x)dx \right ]\\ &=2\left ( \arcsin \left ( \dfrac{t}{4} \right ) \right )_0^2\\ &=2\left ( \arcsin \left ( \dfrac{2}{4} \right )-\arcsin \left ( \dfrac{0}{4} \right ) \right )\\ &=2\left ( \arcsin \left ( \dfrac{1}{2} \right )-0 \right )\\ &=2\left ( \dfrac{\pi}{6} \right )\\ &=\dfrac{\pi}{3} \end{align} {/eq}


{eq}\color{blue}{\boxed{\int_1^3 \dfrac{2}{\sqrt{16 - (x-1)^2}} \,dx=\dfrac{\pi}{3}}} {/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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