Definite Integral

The definite integral of $ƒ(x)$ from $a$ to $b$ is defined by the expression \[\int_a^b ƒ(x)dx\] Here, a is said to be the lower limit and b, the upper limit. The definite integral has a definite value. If $ƒ$ is a continuous function and $F(x)=\int_a^b ƒ(t)dt$, then, \[\frac{d}{dx}F(x)=ƒ(x)\] This theorem is known as the Fundamental Theorem of Integral Calculus. Now with the help of this fundamental theorem, we can prove the following corollary.

Corollary: If $ƒ$ is continuous on $[a,b]$ and $\phi$ is an antiderivative of $ƒ$, then \[\int_a^b ƒ(x)dx=\phi(b)-\phi(a)\]

Proof: Let \[F(x)=\int_a^x ƒ(t)dt\] Obviously, we get $F(a)=0$. Now, as $F$ and $\phi$ are the antiderivatives of the same function $ƒ$, they differ only by a constant. So \[F(x)=\phi(x)+C\] \[\therefore F(a)=\phi(a)+C\] \[\text{or, }0=\phi(a)+C\] \[\text{or, }\phi(a)=-C\] \[\therefore F(x)=\phi(x)-\phi(a)\] \[\text{and, }F(b)=\phi(b)-\phi(a)\] But we have \[F(b)=\int_a^b ƒ(t)dt\] \[\therefore \int_a^b ƒ(t)dt=\phi(b)-\phi(a)\] This theorem is known as the Fundamental Theorem of Integral Calculus.

Change of Limits

Sometime we use substitution method to evaluate a definite integral. If we put $ƒ(x)=y$, the variable $x$ has been changed into the variable $y$. But in a definite integral $\int_a^b ƒ(x)dx$, the limits are of the variable $x$, not of $y$. So, while evaluating a definite integral by substitution method, the limits of $x$ must be changed into the limits of $y$.

Evaluate:

$\int_1^2(2x^2+3x+4)dx$

\[\int(2x^2+3x+4)dx=\frac{2}{3}x^3+\frac{3}{2}x^2+4x\] \[\therefore \int_1^2(2x^2+3x+4)dx=\frac{2}{3}[x^3]_1^2+\frac{3}{2}[x^2]_1^2+4[x]_1^2\] \[=\frac{2}{3}(8-1)+\frac{3}{2}(4-1)+4(2-1)=\frac{79}{6}\]


$\int_0^1\frac{2xdx}{x^2+3}$

\[\text{Let }I=\int\frac{2xdx}{x^2+3}\] Put $x^2+3=t$, then $2xdx=dt$. \[\therefore I=\int\frac{dt}{t}=\log t=\log(x^2+3)\] \[\therefore \int\frac{2xdx}{x^2+3}=\log(x^2+3)\] \[\therefore\int_0^1\frac{2xdx}{x^2+3}=[\log(x^2+3)]_0^1\] \[=\log(1+3)-\log(0+3)\] \[=\log 4-\log 3\]


$\int_0^{\frac{π}{4}}\frac{dx}{1-\sin x}$

\[\int\frac{dx}{1-\sin x}=\int\frac{1+\sin x}{1-\sin^2x}dx\] \[=\int\frac{1+\sin x}{\cos^2x}dx\] \[=\int\sec^2dx+\int\tan x\sec xdx\] \[=\tan x+\sec x\] \[\therefore \int_0^{\frac{π}{4}} \frac{dx}{1-\sin x}=[\tan x]_0^{\frac{π}{4}}+[\sec x]_0^{\frac{π}{4}}\] \[=(\tan\frac{π}{4}-\tan 0)+(\sec\frac{π}{4}-\sec 0)\] \[=(1-0)+\sqrt{2}-1\] \[=\sqrt{2}\]


$\int_1^e x\log xdx$

\[\int x\log xdx\] \[=\log x\int xdx-\int\left(\frac{d}{dx}(\log x)\int xdx\right)dx\] \[=\log x.\frac{x^2}{2}-\frac{1}{2}\int xdx\] \[=\frac{1}{2}x^2\log x-\frac{1}{4}x^2\] \[\int_1^e x\log xdx=\frac{1}{2}[x^2\log x]_1^e-\frac{1}{4}[x^2]_1^e\] \[=\frac{1}{2}(e^2\log e-\log 1)-\frac{1}{4}(e^2-1)\] \[=\frac{1}{2}(e^2-0)-\frac{e^2}{4}+\frac{1}{4}\] \[=\frac{e^2}{4}+\frac{1}{4}=\frac{1}{4}(e^2+1)\]


$\int_1^2\frac{\sin(\log t)}{t}dt$

\[\int_1^2\frac{\sin(\log t)}{t}dt\] Put $y=\log t$, then, $dy=\frac{1}{t}dt$. When $t=1$, $y=0$ and when $t=2$, $y=\log 2$. \[\therefore \int_1^2\frac{\sin(\log t)}{t}dt=\int_0^{\log 2}\sin ydy=[-\cos y]_0^{\log 2}\] \[=1-\cos(\log 2)\]


$\int_0^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^2}}$

\[\text{Let }I=\int\frac{dx}{\sqrt{1-x^2}}\] Put $x=\sin\theta$, then, $dx=\cos\theta d\theta$. \[I=\int\frac{\cos\theta d\theta}{\sqrt{1-\sin^2\theta}}=\int 1d\theta=\theta\] \[\therefore \int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x\] \[\therefore \int_0^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^2}}=[\sin^{-1}x]_0^{\frac{1}{2}}\] \[=\sin^{-1}\left(\frac{1}{2}\right)-\sin^{-1}(0)=\frac{π}{6}-0=\frac{π}{6}\]


$\int_0^a\frac{dx}{a^2+x^2}$

\[\text{Let }I=\int\frac{dx}{a^2+x^2}\] Put $x=a\tan\theta$, then, $dx=a\sec^2\theta d\theta$. \[\therefore I=\int\frac{a\sec^2\theta}{a^2\sec^2\theta}=\frac{1}{a}\int 1d\theta =\frac{1}{a}\theta\]\[=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)\] \[\therefore \int_0^a\frac{dx}{a^2+x^2}=\frac{1}{a}\left[\tan^{-1}\left(\frac{x}{a}\right)\right]_0^a\] \[=\frac{1}{a}[\tan^{-1}(1)-\tan^{-1}(0)]=\frac{1}{a}\left[\frac{π}{4}-0\right]=\frac{π}{4a}\]

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