Learn fundamental theorem of calculus 1 & 2 with chain rule in combination, practice problems on use of first & second fundamental theorem of calculus.

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.

A \(\dfrac {-5}{6}\)

B \(\dfrac {7}{2}\)

C \(\dfrac {18}{91}\)

D 27

- 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.

A \((3+x^3)^4\)

B \((2+x^7)^5\)

C \((5-x)^2\)

D \(\dfrac {7}{x}\)

- 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.

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

A \(2x^3\,cos\,x^2+(6-9x)\,cos(2-3x)\)

B \(2x\,sin\,x^2+4x^3\,cos\,x\)

C \(4x\,sin\,x+x^2\,(cos\,x)\)

D \(cos(7-8x)\,sin\,x^2\)

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

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

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

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

A 288

B \(\dfrac {-1}{7}\)

C –132

D \(\dfrac {432}{5}\)

- 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 \(x\).

A \(\dfrac {x^{14}+1}{x^{28}+1}×7x^6\)

B \((x^2+1)×85x^2\)

C \(\dfrac {12}{x^{2}+7}\)

D \(5\,sin\,x\)

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 \(C=0\).

A \(\dfrac {41}{8}\)

B \(\dfrac {56}{3}\)

C \(\dfrac {2}{9}\)

D \(\dfrac {7}{18}\)