Antiderivatives

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)$ $\therefore \int_0^a\frac{dx}{a^2+x^2}=\frac{1}{a}\left[\tan^{-1}\left(\frac{x}{a}\right)\right]_0^a$ $=\frac{1}{a}[\tan^{-1}(1)-\tan^{-1}(0)]=\frac{1}{a}\left[\frac{π}{4}-0\right]=\frac{π}{4a}$