Subject

# Fundamental Theorem of Calculus

## Function expressed through Integral having Variable as Upper Limit

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.

Illustration Question

## Fundamental Theorem of Calculus Part 1 (FTC 1)

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

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

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

= $$f(u)Ã—\dfrac {du}{dx}$$                             (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

## Fundamental Theorem of Calculus 2 (FTC 2)

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

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

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

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

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