Definite Integration Set 6

Example 11

Evaluate these limits:

(a) \(\begin{align}{L_1} = \mathop {\lim }\limits_{x \to 0} \frac{{\int\limits_0^x {(1 - \cos 2x)\,dx} }}{{x\int\limits_0^x {\tan x\,dx} }}\end{align}\)         (b) \(\begin{align}{L_2} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {\int\limits_0^x {{e^x}dx} } \right)}^2}}}{{\int\limits_0^x {{e^{{x^2}}}dx} }}\end{align}\)

Solution: Observe that both these limits are of the from \(\begin{align}\frac{0}{0}\end{align}\) , and therefore, we can use the L.H. rule. To differentiate the integrals, we can use the Leibnitz’s differentiation rule.

\(\begin{align}(a) \qquad \qquad &{L_1} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos 2x}}{{x\tan x + \int\limits_0^x {\tan x\,dx} }} \left( {{\rm{still\;of\;the\;form }}\frac{0}{0}} \right)\\&\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{2\sin 2x}}{{x{{\sec }^2}x + \tan x + \tan x}}\\\,\,\,\,\,\, &\;\;\;= \mathop {\lim }\limits_{x \to 0} \frac{{4\sin x\cos x}}{{x{{\sec }^2}x + 2\tan x}}\\\,\,\,\,\,\, &\;\;\;= \mathop {\lim }\limits_{x \to 0} \frac{{4\sin x{{\cos }^3}x}}{{x + 2\sin x\cos x}}\\\,\,\,\,\,\, &\;\;\;= \mathop {\lim }\limits_{x \to 0} \frac{{4{{\cos }^3}x}}{{\left( {\frac{x}{{\sin x}}} \right) + 2\cos x}}\\\,\,\,\,\,\, &\;\;\;= \frac{4}{{1 + 2}}\\\,\,\,\,\,\, &\;\;\;= \frac{4}{3}\end{align}\)

\(\begin{align}\text{(b)}\qquad\qquad& {L_2} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\int\limits_0^x {{e^x}dx} } \right).{e^x}}}{{x \cdot {e^{{x^4}}}}} \,\,\left( {{\text{still of the form }}\frac{0}{0}} \right)\\&\quad= \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{e^x}}}{{{e^{{x^4}}}}}} \right) \cdot \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\int\limits_0^x {{e^x}dx} }}{x}} \right)\left\{ \begin{gathered}{\rm{The\; second\; limit\; is\; of }}\\{\rm{the\; form\; }}\frac{0}{0}\end{gathered} \right\}\\&\quad = 1 \cdot \mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{1}\\&\quad= 1\end{align}\)

Example -12

Find a function \(g(x)\) continuous in \((0,\infty )\) such that \(g(x) > 0\,\,\,\,\rlap{-} V\,\,\,\,x \in (0,\infty )\) and \(g(0) = 0,\) and

\[\int\limits_0^x {{g^2}(t)dt = \frac{2}{x}{{\left( {\int\limits_0^x {g(t)dt} } \right)}^2}} \]

 Solution: Differentiating the given relation w.r.t. x, we obtain:

\[{g^2}(x) = \frac{4}{x}\left( {\int\limits_0^x {g(t)dt} } \right) \cdot g(x) - \frac{2}{{{x^2}}}{\left( {\int\limits_0^x {g(t)dt} } \right)^2}\]

Let \(\int\limits_0^x {g(t)dt = y.} \) Thus,

\[\begin{align}&\quad\,\,\,\,\,\,\,{g^2}(x) = \frac{{4y\,g(x)}}{x} - \frac{{2{y^2}}}{{{x^2}}}\\ &\Rightarrow\quad 2{y^2} - 4xg(x)y + {x^2}{g^2}(x) = 0\\ &\Rightarrow\quad y = \frac{{4xg(x) \pm \sqrt {16{x^2}{g^2}(x) - 8{x^2}{g^2}(x)} }}{4}\\\,\,\,\, &\qquad\;\;\;\;= \frac{{\left( {4 \pm 2\sqrt 2 } \right)xg(x)}}{4}\\\,\,\,\, &\qquad\;\;\;\;= \left( {1 \pm \frac{1}{{\sqrt 2 }}} \right)x\,\,g(x)\end{align}\]

Thus,

\[\int\limits_0^x {g(t)dt = kx\,g(x) \left\{ {k = 1 \pm \frac{1}{{\sqrt 2 }}} \right\}} \]

Differentiating both sides, we obtain:

\[\begin{align}&\quad\,\,\,\,\,\,\,\,g(x) = k\left\{ {g(x) + xg'(x)} \right\}\\ &\Rightarrow\quad (1 - k)g(x) = \frac{{kx\,d\,g(x)}}{{dx}}\\ &\Rightarrow\quad \frac{{dg(x)}}{{g(x)}} = \left( {\frac{{1 - k}}{k}} \right)\frac{{dx}}{x}\end{align}\]

Integrating both sides, we obtain,

\[\begin{align}&\text{ln}\;g(x) = \left( {\frac{{1 - k}}{k}} \right)\ln x + C\\\\\Rightarrow \qquad&g(x) = {C_0}{x^{\frac{{1 - k}}{k}}} \qquad\qquad\left( \begin{array}{l}{\rm{Take\;}}C{\;\rm{\;as\;ln\;}}{C_0}{\rm{\;where\;}}\\{C_0}{\rm{\;is\;another\; constant}}\end{array} \right)\end{align}\]

Since \(g(0) = 0,\) the power of x must be positive, i.e.,

\[k = 1 - \frac{1}{{\sqrt 2 }}\]

Thus,

\[g(x) = {C_0}{x^{\frac{1}{{\sqrt 2 - 1}}}} = {C_0}{x^{\sqrt 2 + 1}}\]