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Consider the definite integral \int_6^{12} x(\sqrt{(12x-x^2)} dx. Then the most appropriate...

Question:

Consider the definite integral {eq}\int_6^{12} x\sqrt{(12x-x^2)} dx {/eq}.

Then the most appropriate substitution to simplify this integral is x = g(t) where g(t) = ?

After making this substitution and simplifying (using trig identities), we obtain the integral {eq}\int_a^b f(t)dt {/eq} where f(t) = ? a = ? b = ? After evaluating this integral we obtain: {eq}\int_6^{12} x(\sqrt {(12x-x^2)} dx {/eq}= ?

Integration by Substitution:

There are two types of substitution: u or v substitution and the trigonometric substitution. In this problem, we used both methods.

U-substitution: It can be applied when integral is complicated to solve using direct methods.

Trigonometric substitution: It can be applied when we need a trigonometric function for solving the integral.


Basic integration formulas:

1. Complete the square: {eq}\displaystyle x^2+2ax+a^2=\left(x+a\right)^2. {/eq}

2. The sum rule: {eq}\displaystyle \int f\left(r\right)\pm g\left(r\right)dr=\int f\left(r\right)dr\pm \int g\left(r\right)dr. {/eq}

3. Change the limits: {eq}\displaystyle \int _a^bf\left(x\right)dx=-\int _b^af\left(x\right)dx, a<b. {/eq}

4. Remove the constant from the integration: {eq}\displaystyle \int a\cdot f\left(r\right)dr=a\cdot \int f\left(r\right)dr. {/eq}

5. The radical rule: {eq}\displaystyle \sqrt{a}=a^{\frac{1}{2}}. {/eq}

6. The power rule: {eq}\displaystyle \int r^adr=\frac{r^{a+1}}{a+1}, \quad a\ne -1. {/eq}

7. The trigonometric identity: {eq}\displaystyle \cos ^2\left(x\right)=\frac{1+\cos \left(2x\right)}{2}. {/eq}

8. Integration of a constant: {eq}\displaystyle \int adr=ar. {/eq}

9. Common integration: {eq}\displaystyle \int \cos \left(w\right)dw=\sin \left(w\right). {/eq}

Answer and Explanation:

We have to solve the integration of $$\displaystyle I = \int_6^{12} x\sqrt{(12x-x^2)} dx $$

Complete the square.

$$\begin{align*} \displaystyle &= \int_6^{12} x\sqrt{-\left(x^2-12x+\left(-6\right)^2-\left(-6\right)^2\right)} \ dx \\ \displaystyle &=\int_6^{12} x\sqrt{-\left(\left(x-6\right)^2-\left(-6\right)^2\right)} \ dx\\ \displaystyle &= \int _6^{12}x\sqrt{-\left(x-6\right)^2+36} \ dx \end{align*} $$

Apply the substitution for {eq}u=x-6 \Rightarrow du = dx. {/eq}

Limits: {eq}6 \rightarrow 0 {/eq} and {eq}12 \rightarrow 6. {/eq}

$$\displaystyle = \int _0^6\left(u+6\right)\sqrt{-u^2+36} \ du $$

Expand:

$$\displaystyle = \int _0^6u\sqrt{-u^2+36}+6\sqrt{-u^2+36} \ du $$

Apply the sum rule.

$$\begin{align*} \displaystyle &= \int _0^6u\sqrt{-u^2+36}du+\int _0^66\sqrt{-u^2+36} \ du \\ \displaystyle &= I_1+I_2 \end{align*} $$

Now, we are going to find {eq}I_1. {/eq}

{eq}\displaystyle I_1 = \int _0^6u\sqrt{-u^2+36} \ du \\ {/eq}

Apply the substitution for {eq}v=-u^2+36 \Rightarrow dv = -2u du. \\ {/eq}

Limits: {eq}0 \rightarrow 36 {/eq} and {eq}6 \rightarrow 0. {/eq}

{eq}\displaystyle = \int _{36}^0-\frac{\sqrt{v}}{2} \ dv \\ {/eq}

Change the limits.

{eq}\displaystyle = -\int _0^{36}-\frac{\sqrt{v}}{2} \ dv \\ {/eq}

Remove the constant from the integration.

{eq}\displaystyle = -\left(-\frac{1}{2}\cdot \int _0^{36}\sqrt{v}dv\right) \\ {/eq}

Apply the radical rule.

{eq}\displaystyle = -\left(-\frac{1}{2}\cdot \int _0^{36}v^{\frac{1}{2}}dv\right) \\ {/eq}

Apply the power rule.

{eq}\displaystyle = \frac{1}{2}\left[\frac{2v^{\frac{3}{2}}}{3}\right]^{36}_0 \\ {/eq}

Compute the boundaries.

{eq}\begin{align*} \displaystyle &= \frac{1}{2}\left[144-0\right]\\ \displaystyle &= 72. \end{align*} \\ {/eq}

Now, solving for {eq}I_2. {/eq}

{eq}\displaystyle I_2 = \int _0^66\sqrt{-u^2+36} \ du {/eq}

Remove the constant from the integration.

{eq}\displaystyle = 6\cdot \int _0^6\sqrt{-u^2+36} \ du \\ {/eq}

Apply the trigonometric substitution for {eq}u=6\sin \left(v\right) \Rightarrow du=6\cos \left(v\right) dv. \\ {/eq}

Limits: {eq}0 \rightarrow 0 {/eq} and {eq}6 \rightarrow \frac{\pi }{2}. {/eq}

{eq}\displaystyle = 6\cdot \int _0^{\frac{\pi }{2}}36\cos ^2\left(v\right) \ dv \\ {/eq}

Remove the constant from the integration.

{eq}\displaystyle = 6\cdot 36\cdot \int _0^{\frac{\pi }{2}}\cos ^2\left(v\right) \ dv \\ {/eq}

Use the trigonometric identity.

{eq}\displaystyle = 6\cdot 36\cdot \int _0^{\frac{\pi }{2}}\frac{1+\cos \left(2v\right)}{2} \ dv \\ {/eq}

Remove the constant from the integration.

{eq}\displaystyle = 6\cdot 36\cdot \frac{1}{2}\cdot \int _0^{\frac{\pi }{2}}1+\cos \left(2v\right) \ dv \\ {/eq}

Apply the sum rule.

{eq}\displaystyle = 6\cdot 36\cdot \frac{1}{2}\left(\int _0^{\frac{\pi }{2}}1dv+\int _0^{\frac{\pi }{2}}\cos \left(2v\right)dv\right) \\ {/eq}

Apply the substitution for {eq}w=2v \Rightarrow dw=2 dv. {/eq}

Limits: {eq}0 \rightarrow 0 {/eq} and {eq}\frac{\pi }{2} \rightarrow \pi. {/eq}

{eq}\displaystyle = 6\cdot 36\cdot \frac{1}{2}\left(\int _0^{\frac{\pi }{2}}1dv+\int _0^{\pi }\cos \left(w\right)\frac{1}{2}dw\right) \\ {/eq}

Remove the constant from the integration.

{eq}\displaystyle = 6\cdot 36\cdot \frac{1}{2}\left(\int _0^{\frac{\pi }{2}}1dv+\frac{1}{2}\cdot \int _0^{\pi }\cos \left(w\right)dw\right) \\ {/eq}

Use the common integration.

{eq}\displaystyle = 6\cdot 36\cdot \frac{1}{2}\left(\left[v\right]^{\frac{\pi }{2}}_0+\frac{1}{2}\left[\sin \left(w\right)\right]^{\pi }_0\right) \\ {/eq}

Compute the boundaries.

{eq}\begin{align*} \displaystyle &= 6\cdot 36\cdot \frac{1}{2}\left(\left[\frac{\pi }{2}-0\right]+\frac{1}{2}\left[0-0\right]\right)\\ \displaystyle &= 6\cdot 36\cdot \frac{1}{2}\left(\frac{\pi }{2}+0\right)\\ \displaystyle &= 54\pi. \end{align*} \\ {/eq}

Put the value of {eq}I_1 {/eq} and {eq}I_2 {/eq} in {eq}I. {/eq}

$$\displaystyle I = 72+54\pi . $$


Learn more about this topic:

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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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