4.4 Area by Definite Integrations

Rule of Method of Substitution

If \( u = g(x) \) is a differentiable function whose range is an interval I , and f is continuous on I , then \[ \int_{a}^b g'(x) f(g(x)) dx = \int_{g(a)}^{g(b)} f(u) du. \]

• Substitution Method for Definite Integration   Problems   Q1 ; Q2 ; Q3 ; Q4 ; Q5 ; Q6 ;
• Area Between two curves

Solving definite integrations by Method of Substitution

Q. Evaluate \(\int_{0}^2 r^2\sqrt{8-r^3} dr \).

Ans. The integration can be solved by two methods

Method 1	Method 2
In this method, we will solve the integration as indefinite integration by substitution method, and then using fundamental theorem of calcululs, we will get the value of the integartion.	In this method, we substitute the require expression by a new variable, and change the limit for new variable. Then solve the integration using fundamental theorem on calculus.
Put $t= 8- r^3$. Then \[dt= – 3 r^2 dr \implies r^2 dr = – \frac{dt}{3} \]	Put $t= 8- r^3$. Then \[dt= – 3 r^2 dr \implies r^2 dr = – \frac{dt}{3} \]
The indefinite form of the given integration reduces to \[- \int \sqrt{t} \frac{dt}{3} = – \frac{1}{3}. \frac{t^{3/2}} {3/2} = – \frac{2}{9} (8-r^3)^{3/2}\]	Now, $r=0 \implies t=8-0^3=8$ and $r=2 \implies t=8-2^3=0$
Finally, by fundamental theorem, we have \[\begin{align*} \int_0^2 r^2\sqrt{8-r^3} dr &= – \frac{2}{9} \left((8-r^3)^{3/2} \right)_{0}^2\\&= – \frac{2}{9}( (8-8)^{3/2} – 8^{3/2})\\ &= \frac{32\sqrt{2}}{9}. \end{align*}\]	The given definite integration reduces to \[ – \int_8^0 \sqrt{t} \frac{dt}{3} = – \frac{2}{9} \left(t^{3/2}\right)_8^0 \]

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Q. Evaluate \(\int_{0}^{\pi/4} \tan^2(t) \sec^2(t) dr \).

Ans. This problem can also be solve by two method as mention in earlier problem.

Method 1	Method 2
Put \(\theta=\tan(t)\) This implies \(d\theta=\sec^2(t) dt\)	Put \(\theta=\tan(t)\) This implies \(d\theta=\sec^2(t) dt\)
The indefinite form of the given definite integration reduces to \[ \int \theta^2 d\theta = \frac{\theta^3}{3}=\frac{1}{3} \tan^3(t).\]	Now, \[t=0 \implies \theta=\tan(0)=0\] and \[t=\pi/4 \implies \theta=\tan(\pi/4)=1.\]
Finally, the given definite integartion can be solve as \[ \begin{align*}&\int_0^{\pi/4} \tan^2(t)\sec^2(t) dt \\ &=\frac{1}{3} \bigg[ \tan^3(t)\bigg]_0^{\pi/4}\\&=\frac{1}{3} \bigg[ \tan^3(\pi/4)-\tan^3(0) \bigg]\\&=\frac{1}{3}\left(1-0\right) =-\frac{1}{3}\end{align*}.\]	Thus, the given definite integartion reduces to \[ \begin{align*}\int_0^{\pi/4} \tan^2(t)\sec^2(t)dt &= \int_0^1 \theta^2 d\theta \\&= \frac{1}{3} \left(\theta^3\right)_0^{1}\\ & =\frac{1}{3} \left(1-0\right)=\frac{1}{3}.\end{align*}\]

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Q. Evaluate \(\int_{-2}^{2} \frac{x^2}{(9+x^3)^2} dx \).

Ans. This problem also can be solved by two methods. But we will solve this by Method 2 only.

• Put $ t = 9+x^3$. This implies $dt= 3 x^2 dx \implies x^2 dx = \frac{dt}{3} $
• Now, \[ \begin{align*}&x=-2 \implies t= 9+ (-2)^3=9-8=1; and &x=2 \implies t = 9 + (2)^3=9+8=17.\end{align*} \]
• The given definite integartion reduces to \[ \frac{1}{3}\int_1^{17} \frac{1}{t^2} dt= \frac{1}{3} \left(-\frac{1}{t}\right)_{1}^{17} = -\frac{1}{3}\left(1- \frac{1}{17}\right) = – \frac{16}{51}. \]

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Q. Evaluate \(\int_{0}^{2\pi} \frac{\cos(2x)}{\sqrt{3+4\sin(2x)}} dx \).

Ans. This problem also can be solved by two methods. But we will solve this by Method 2 only.

• Put $ t = 3+4 \sin(2 x)$. This implies $dt= 8 \cos(2x)dx $
• Now
• Now, \[ \begin{align*}&x=0 \implies t= 3+ 4 \sin(0)=3; and &x=2\pi \implies t=3+ 4 \sin(4\pi)=3.\end{align*} \]
• The given definite integartion reduces to \[ \frac{1}{8}\int_3^{3} \frac{1}{\sqrt{t}} dt= 0. \]

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Q. Evaluate \(\int_{0}^{\pi/2} \frac{\cos(t)}{1+\sin^2(t)} dt \).

Ans. This problem also can be solved by two methods. But we will solve this by Method 2 only.

• Put $ y = \sin(t)$. This implies $dy= \cos(t)dt $
• Now, \[ \begin{align*}&t=0 \implies y= \sin(0)=0; and &t=\pi/2 \implies y=\sin(\pi/2)=1.\end{align*} \]
• The given definite integartion reduces to \[ \int_0^{1} \frac{1}{1+y^2} dy= \left(\tan^{-1}(y)\right)_{0}^1 =\tan^{-1}(1)-\tan^{-1}(0) =\pi/4. \]

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Q. Evaluate \(\int_{0}^{3} \frac{1}{\sqrt{9-r^2}} dr \).

Ans. This problem also can be solved by two methods. But we will solve this by Method 2 only.

• Put $ r = 3 \sin(t)$. This implies $dr= 3\cos(t)dt $
• Now, \[ \begin{align*}&r=0 \implies 3 \sin(t)=0 \implies t=0; \\ and &r=3 \implies 3 \sin(t)=3 \implies \sin(t)=1 \implies t=\pi/2.\end{align*} \]
• The given definite integartion reduces to \[\begin{align*}\int_0^{\pi/2} \frac{3 \cos(t)}{\sqrt{9-9\sin^2(t)}}dt &= \int_0^{\pi/2} \frac{ \cos(t)}{\sqrt{1-\sin^2(t)}} dt\\&= \int_0^{\pi/2} \frac{ \cos(t)}{\cos(t)} dt \\&=\int_0^{\pi/2} dt= (t)_0^{\pi/2} =\pi/2. \end{align*} \]

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Q. Evaluate \(\int_{\ln(1/2)}^0 \frac{e^x}{\sqrt{1-e^{2x}}} dx \).

Ans. This problem also can be solved by two methods. But we will solve this by Method 2 only.

• Put $ t= e^x $. This implies $dt= e^x dx $
• Now, \[ \begin{align*}&x=0 \implies t=e^0=1; \\ and \; &x= \ln(1/2) \implies t=e^{\ln(1/2)}=1/2.\end{align*} \]
• The given definite integartion reduces to \[\begin{align*}\int_{1/2}^{1} \frac{1}{\sqrt{1-t^{2}}}dt &= \left(\sin^{-1}(t)\right)_{1/2}^{1}\\ &= \sin^{-1}(1)- \sin^{-1}(1/2)=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}. \end{align*} \]

Area of a region by definite integrations

One of the most important geometric nature of the definite integration is that it will provide us the region bounded by the functions $y=f(x)$ and the lines $x=a$ and $x=b$. We will explain the method by examples. There are two kind of area problems.

• Area bounded by a curve y=f(x), x-axis, lines $x=a$ and $x=b$. The problem need a slight different approach if $f(x)$ cross the $x$-axis in the interval $(a, b)$ .
• Area bounded by two curves y=f(x), y=g(x), lines $x=a$ and $x=b$.
  Consider two curve $y=f(x)$ and $y=g(x)$ such that they intersect at $x=a$ and $x=b$, and $g(x) \geq f(x) $. Then, the area between $y=f(x)$ and $y=g(x)$ is given by \[\int_{a}^b \bigg(g(x) – f (x) \bigg) dx.\]

We will solve both kind of problems.

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Q. Find the area of the region bounded by $y=\frac{1}{x+3}$, and the line $x=-2$ and $x=2$.

Ans. The function $\frac{1}{x+3}$ never cross the $x$-axis. Hence, the area is given by direct integration from $x=-2$ to $x=2$. Thus, \[ \begin{align*} A= \int_{-2}^{2} \frac{1}{x+3}dx = \left(\ln(x+3)\right)_{-2}^2= \ln(5)-\ln(1)=\ln(5) \end{align*}\]

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Q. Find the area of the region above $y=x^2$, and below $y=2 – x^2$.

Ans. To find the area of the Region , between the given curves $y=x^2$ and $y=2 – x^2$, first we need to find the points of intersections of the curves. At the point of intersection, we have \[ \textcolor{blue}{x^2 = 2- x^2} \implies \textcolor{blue}{2 x^2 = 2} \implies \textcolor{blue}{x^2=1} \implies \textcolor{blue}{\fbox{$x= \pm 1$}}. \] \[ \begin{align*} A= \int_{-1}^{1} \left( (2- x^2) – x^2\right) dx = \left( 2 x – \frac{2}{3}x^3\right)_{-1}^1 = \left( 2 – \frac{2}{3}\right)-\left( -2 + \frac{2}{3}\right)= \frac{4}{3}+ \frac{4}{3} =\frac{8}{3} \end{align*}\]

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Remark : Few common questions by the students, do we need to draw the graph? How we know which function is bigger?
The clarification about those questions are as follows:

• It is always good to know the graph of a curve that makes the solution much easier. However, unless the question ask, it is not necessary to draw the graph.
• If you can draw the graph, then it is easy to get upper curve and lower curve. In other case, read the statement of questions carefully. For example, in the above question, it is written that above $y=x^2$ and below $y=2-x^2$ which clearly implies $y=2-x^2$ is bigger than $y=x^2$ in the requested region.
• In the case, if you are completely uncertain, then chose any one curve as $g$ and other one as $f$ to have $g – f$. If you have wrong selection, then the answer will come negative.

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