Subject

Fundamental Theorem of Calculus

Function expressed through Integral having Variable as Upper Limitlogo

Consider

\(g(x)=\int\limits_a^xf(t)\;dt\) where \('f' \)is a continuous function and \(x \) varies between \('a'\) and \('b'\). If \(x\) varies, then this number varies and hence \(g(x)\) is a function of \(x\).

\(g(x)\) indicates a variable area whose lower limit is fixed and upper limit is variable.

Function expressed through Integral having Variable as
Illustration Question

Consider the graph of a function \('f'\) as shown.

Let \(g(x)=\int\limits_0^xf(t)\,dt\). The value of \(g(2)\) is 

Consider the graph of a function \('f'\) as shown.

Let \(g(x)=\int\limits_0^xf(t)\,dt\).
 

Fundamental Theorem of Calculus Part 1 (FTC 1)logo

If \('f'\) is a continuous function on [a, b] then function \('g'\) defined by

\(g(x)=\int\limits_a^xf(t)\,dt\) \(a\leq x\leq b\) is continuous on [a, b] and differentiable on (a, b) and \(g'(x)=f(x)\).

In short, if \(g(x)=\int\limits_a^x\,f(t)\;dt \Rightarrow g'(x)=f(x)\) 

\(\dfrac {d}{dx} \left (\int\limits_a^x\,f(t)\;dt\right)=f(x)\rightarrow\)Leibnitz notation for derivatives.

If we integrate a function  \('f'\) and then differentiate it, we get the same function back.

Illustration Question

If \(g(x)=\int\limits_1^x\,(3+t^3)^4\;dt \), then  \( g'(x) \) will have the expression

 

Chain Rule in Combination with FTC 1(only when upper limit is a function of \(x\))logo

If \(g(x)=\int\limits_a^{u(x)}\;f(t)\;dt\), then \(\dfrac {d}{dx}g(x)=\dfrac {d}{du} \left ( \int\limits_a^u\;f(t)\;dt \right)×\dfrac {du}{dx}\rightarrow\) Chain Rule

 = \(f(u)×\dfrac {du}{dx}\)                             (Put  \(u\) in place of t  everywhere)

First put the upper limit in the integrand expression in place of \(x\).

We are assuming lower limit 'a' to be a constant and independent of \(x\).

Illustration Question

If \(g(x)=\displaystyle\int\limits_5^{x^7}\left (\dfrac {t^2+1}{t^4+2}\right)\;dt\) then expression for \(g'(x)\) or \(\dfrac {d}{dx}\,g(x)\) will be 

 

Fundamental Theorem of Calculus 2 (FTC 2)logo

If \('f'\)is continuous in [a, b] then

\(\int\limits_a^b\,f(x)\;dx=F(b)-F(a)=F(x)\Big]_a^b\)

where \(F(x)\) is any anti derivative of \(f\)that is \(F'=f\)

We observe that the  complicated procedure involved in evaluating the definite integral (summation and application of limits) is now replaced by just evaluation of anti derivative at the ends, \(x=a\) and \(x=b\).

\(\int\,x^n\;dx=\dfrac {x^{n+1}}{n+1}+C\) for all values of \(x\).

To evaluate \(\int\limits_a^b\;f(x)\;dx\)

(1) Find \(\int\limits\;f(x)\;dx=F(x)\) (say)

(2) Find \(F(b)-F(a)\)

(3) This is the value of the required definite integral.

\(F(x)\Bigg]_a^b\) is the notation for \(F(b)-F(a)\).

We always take the anti derivative \(F\) in which \(C=0\).

Illustration Question

Find the value of the integral

\(\int\limits_{-3}^1\,(2x^2)\;dx\).

 

Chain Rule in Combination with FTC 1 (when both lower and upper limits are function of x)logo

If \(g(x)=\int\limits_{\phi(x)}^{h(x)}f(t)\;dt\) (both limits are function of \(x\))

Then \(g(x)=\int\limits_{\phi(x)}^{0}f(t)\;dt+ \int\limits_{0}^{h(x)}f(t)\;dt\)

\(=-\int\limits_0^{\phi(x)}\,f(t)dt+\int\limits_0^{h(x)}\,f(t)\;dt\)

Now we find \(g'(x)\) by chain rule.

\(g'(x)=-f(\phi(x))\,\phi'(x)+f(h(x))\,h'(x)\)

\(\Rightarrow g'(x)=f\underbrace {(h(x))}_{\text {upper limit }}\, \;\;\underbrace {h'(x)}_{\text {dev. of U.L.}}-f\underbrace{(\phi(x))}_{\text {lower limit}}\,\;\;\underbrace {\phi'(x)}_{dev. of L.L.}\)

Illustration Question

If \(g(x)=\int\limits_{2-3x}^{x^2}\,(t\,cost)\;dt\), then \(g'(x)\) will have the expression

 

Use of Fundamental Theorem of Calculus 2 (FTC 2)logo

If \('f'\)is continuous in [a, b] then

\(\int\limits_a^b\,f(x)\;dx=F(b)-F(a)=F(x)\Big]_a^b\)

where \(F(x)\) is any anti derivative of \(f\)that is \(F'=f\)

We observe that the  complicated procedure involved in evaluating the definite integral (summation and application of limits) is now replaced by just evaluation of anti derivative at the ends, \(x=a\) and \(x=b\).

\(\int\,x^n\;dx=\dfrac {x^{n+1}}{n+1}+C\) for all values of \(x\).

To evaluate \(\int\limits_a^b\;f(x)\;dx\)

(1) Find \(\int\limits\;f(x)\;dx=F(x)\) (say)

(2) Find \(F(b)-F(a)\)

(3) This is the value of the required definite integral.

\(F(x)\Bigg]_a^b\) is the notation for \(F(b)-F(a)\).

We always take the anti derivative \(F\) in which \(C=0\).

e.g. \(\displaystyle\int\limits\;\dfrac {x+2}{\sqrt x}\;dx\)

\(=\displaystyle\int\limits\;\dfrac {x}{\sqrt x}+\dfrac {2}{\sqrt x}\;dx\rightarrow\) split into two integral

\(=\displaystyle\int\limits\; x^{1/2}+2x^{-1/2}\;dx\)

\( =\dfrac {x^{3/2}}{3/2}+\dfrac {2x^{1/2}}{1/2}\;+C\)

\(=\dfrac{2}{3}(x)^\dfrac{3}{2}+4(x)^\dfrac{1}{2}+C\)

Illustration Question

Evaluate:

\(I=\displaystyle\int\limits_{4}^9\left (\dfrac{t^2+1}{\sqrt t}\right)\;dt\)

 
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