Evaluate the following integrals: (a) integral_0^1 x^2 square root {6 x + 9} dx. (b) integral x...

Question:

Evaluate the following integrals:

(a) {eq}\displaystyle \int_0^1 x^2 \sqrt {6 x + 9}\ dx {/eq}.

(b) {eq}\displaystyle \int \dfrac x {\sqrt {x + 8}}\ dx {/eq}.

(c) {eq}\displaystyle \int \dfrac {(\ln (x))^6} x\ dx {/eq}.

INTEGRATION BY SUBSTITUTION METHOD

{eq}\displaystyle{\text{This question is from integration and we have to find out the integral of the function and we add constant C after integrating this}\\[10pt] \text{because this is indefinite integral. We will use substitution method to solve this question.}\\[10pt] \text{Rule to evaluate}\int f(x)\quad dx \quad\text{by the substitution}\\[10pt] \text{1. In the integrand put x=g(y), dx = g'(y) dy}\\[10pt] \text{2. Evaluate the resulting integral in y.}\\[10pt] \text{3. Express the result obtained in terms of x.}} {/eq}

{eq}\displaystyle{\text{Definite integral} \int_a^b f(x) \ dx = \begin{vmatrix}F(x) \end{vmatrix}_a^b = F(b) - F(a)\\} {/eq}

Answer and Explanation:


(a) {eq}\displaystyle{ \int_0^1 x^2 \sqrt {6 x + 9}\ dx\\[10pt] \text{Rewrite as}\\[10pt] \int_0^1 x^2 \sqrt {3(2 x + 3)}\ dx\hspace{50pt}\\[10pt] \text{ Substitute u = 2x + 3}\\[10pt] \frac{du}{dx} = 2\implies dx = \frac{du}{2}\\[10pt] \text{ Also, 2x + 3 = u}\\[10pt] 2x = u - 3\\[10pt] x = \frac{u - 3}{2}\\[10pt] x^2 =\left( \frac{u - 3} {2}\right)^2\\[10pt] \text{Now, when x = 0, u = 3}\\[10pt] \text{and, when x = 1, u = 5}\\[10pt] \text{(1) becomes,}\\[10pt] = \int_3^5\left(\frac{u - 3}{2}\right)^2\sqrt{3u}\frac{du}{2}\\[10pt] = \int_3^5\frac{(u - 3)^2}{4}\sqrt{3}\sqrt{x}\frac{du}{2}\\[10pt] = \frac{\sqrt{3}}{8}\left[\int _3^5(u - 3)^2\sqrt{u}\ du\right]\\[10pt] = \frac{\sqrt{3}}{8}\left[\int_3^5( u^2 - 6u +9)\sqrt{u} \ du\right]\hspace{50pt}\left(\text{ using} (a- b)^2 = a^2 - 2ab +b^2\right)\\[10pt] = \frac{\sqrt{3}}{8}\left[\int_3^5 u^\frac{5}{2} - 6u^\frac{3}{2} + 9u^\frac{1}{2} \ du\right]\\[10pt] = \frac{\sqrt{3}}{8}\left[\frac{2u^\frac{7}{2}}{7} - \frac{12u^\frac{5}{2}}{5} + 6u^\frac{3}{2}\right]_3^5\hspace{50pt}\left(\text{Using power rule}\quad\int x^n \ dx = \frac{x^{n + 1}}{n + 1} + C\right)\\[10pt] \text{On applying limits we get,}\\[10pt] = \frac{\sqrt{3}}{8}\left[\left(\frac{2(5)^\frac{7}{2}}{7} - \frac{12(5)^\frac{5}{2}}{5} + \frac{6(5)^\frac{3}{2}}{1}\right) - \left(\frac{2(3)^\frac{7}{2}}{7} - \frac{12(3)^{5}{2}}{5} + \frac{6(3)^\frac{3}{2}}{1}\right)\right]\\[10pt] = \frac{\sqrt{3}}{8}\left[\left(\frac{250\sqrt{5}}{7} -\frac{300\sqrt{5}}{5} + 30\sqrt{5}\right) - \left(\frac{54\sqrt{3}}{7} - \frac{108\sqrt{3}}{5} + 18\sqrt{3}\right)\right]\\[10pt] = \frac{\sqrt{3}}{8}\left[\left(\frac{1250\sqrt{5} - 2100\sqrt{5} + 1050\sqrt{5}}{35}\right) - \left(\frac{270\sqrt{3} - 756\sqrt{3} + 630\sqrt{3}}{35}\right)\right]\\[10pt] = \frac{\sqrt{3}}{8}\left[\left(\frac{200\sqrt{5}}{35}\right) - \left(\frac{144\sqrt{3}}{35}\right)\right]\\[10pt] = \left(\frac{25\sqrt{5}}{35} - \frac{432}{280}\right)\\[10pt] \boxed{= \left(\frac{25\sqrt{5}}{35} - \frac{54}{35}\right)}\\ \\ \text{Note: How to solve}\quad(5)^\frac{7}{2} = \sqrt{5}\sqrt{5}\sqrt{5}\sqrt{5}\sqrt{5}\sqrt{5}\sqrt{5} = 5\times 5\times 5\times \sqrt{5} = 125\sqrt{5}\quad\text{similarly,we can solve others also.}\\} {/eq}

(b) {eq}\displaystyle{ \int \dfrac x {\sqrt {x + 8}}\ dx\hspace{50pt}(1)\\[10pt] \text{ Substitute u = x +8}\\[10pt] \frac{du}{dx} = 1\implies du = dx\\[10pt] \text{Also, u = x +8}\implies x = u- 8\\[10pt] \text{(1) becomes,}\\[10pt] = \int \frac{u -8}{\sqrt{u}} \ du\\[10pt] = \int \frac{u}{\sqrt{u}} \ du - 8\int \frac{1}{\sqrt{u}} \ du\\[10pt] = \int u^\frac{1}{2} \ du - 8\int u^\frac{-1}{2} \ du\\[10pt] = \frac{2u^\frac{3}{2}}{3} - 16u^\frac{1}{2}\hspace{80pt}\left(\text{Using power rule}\quad\int x^n \ dx = \frac{x^{n + 1}}{n + 1} + C\right)\\[10pt] \boxed{= \frac{2(x + 8)^\frac{3}{2}}{3} - 16\sqrt{x + 8} + C}\\} {/eq}.

(c) {eq}\displaystyle{\int \dfrac {(\ln (x))^6} x\ dx\hspace{60pt}(1)\\[10pt] \text{Substitute u = }\ln x\\[10pt] \frac{du}{dx} = \frac{1}{x}\\[10pt] du = \frac{1}{x} \ dx\\[10pt] \text{(1) becomes,}\\[10pt] = \int u^6 \ du\\[10pt] = \frac {u^7}{7} +C\hspace{120pt}\left(\text{Using power rule}\quad\int x^n \ dx = \frac{x^{n + 1}}{n + 1} + C\right)\\[10pt] \boxed{= \frac{(\ln(x))^7}{7} + C}\\} {/eq}.


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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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