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Evaluate the integral:

{eq}\displaystyle \int_0^3 \sqrt {6 x - x^2} \ dx {/eq}.

Question:

Evaluate the integral:

{eq}\displaystyle \int_0^3 \sqrt {6 x - x^2} \ dx {/eq}.

Working rule in Definite Integration:

Let function {eq}m {/eq} is continous on {eq}[p,q] {/eq} and {eq}M {/eq} is a differentiable function on {eq}(p,q) {/eq} such that, {eq}\forall \,x \in (p,q),\,\frac{d}{{dx}}M(x) = m(x) {/eq} then,

{eq}\int\limits_p^q {m(x)dx = M(p) - M(q)} {/eq}.

The integrated function must be evaluated in both the extremes of the interval and subtract their values.

Answer and Explanation: 1

Given that: {eq}\displaystyle \int\limits_0^3 {\sqrt {6x - {x^2}} dx} {/eq}

{eq}\displaystyle\ \eqalign{ & \int\limits_0^3 {\sqrt {6x - {x^2}} dx} = \int\limits_0^3 {\sqrt {9 - 9 + 6x - {x^2}} dx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int\limits_0^3 {\sqrt {9 - {{(x - 3)}^2}} dx} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Factor: }}9 - {{(x - 3)}^2} = 9 - 9 + 6x - {x^2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int\limits_0^3 {\sqrt {{3^2} - {{(x - 3)}^2}} dx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {\frac{{x - 3}}{2}\sqrt {{3^2} - {{(x - 3)}^2}} + \frac{{{3^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{{x - 3}}{3}} \right)} \right]_0^3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\int {\sqrt {{a^2} - {x^2}} dx = \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{x}{a}} \right) + c} } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {\left( {\frac{{3 - 3}}{2}\sqrt {{3^2} - {{(3 - 3)}^2}} + \frac{{{3^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{{3 - 3}}{3}} \right)} \right) - \left( {\frac{{0 - 3}}{2}\sqrt {{3^2} - {{(0 - 3)}^2}} + \frac{{{3^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{{0 - 3}}{3}} \right)} \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {0 - \left( {0 + \frac{9}{2}{{\sin }^{ - 1}}\left( { - 1} \right)} \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{9}{2}{\sin ^{ - 1}}\left( 1 \right) \cr & \int\limits_0^3 {\sqrt {6x - {x^2}} dx} = \frac{{9\pi }}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\sin \frac{\pi }{2} = 1} \right) \cr} {/eq}


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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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