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Fundamental Theorem Of Calculus

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

Consider the graph of a function \('f'\) as shown. Let \(g(x)=\int\limits_0^xf(t)\,dt\). The value of \(g(2)\) is 

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

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

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

D 27

×

\(g(2)=\)Area of shaded portion

image

\(g(2)=\int\limits_0^2 f(t)\;dt\)

(Put \(x=2\))

image

\(= \underbrace {\left(\dfrac {1}{2}×(2+1)×1\right)}_{\text {trapezium OABC}}+ \underbrace {\left(\dfrac {1}{2}×(1+3)×1\right)}_{\text {trapezium ABED}} \)

image

\(=\dfrac {1}{2}×3 + \dfrac {1}{2}×4\)

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

image

Consider the graph of a function \('f'\) as shown. Let \(g(x)=\int\limits_0^xf(t)\,dt\). The value of \(g(2)\) is 

image
A

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

.

B

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

C

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

D

27

Option B is Correct

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 Questions

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

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

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

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

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

×

\(g(x)=\int\limits_1^x(3+t^3)^4\;dt\)

\(\Rightarrow\)\(g'(x)=\)\((3+x^3)^4\) (Just replace \('t'\) by \('x'\)in the integrand expression.) 

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

A

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

.

B

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

C

\((5-x)^2\)

D

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

Option A is Correct

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

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 

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

B \((x^2+1)×85x^2\)

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

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

×

\(g(x)=\displaystyle\int\limits_5^{x^7} \left ( \dfrac {t^2+1}{t^4+2} \right)\;dt\)

\(\Rightarrow \dfrac {d}{dx}\,g(x)= \left ( \dfrac {(x^7)^2+1}{(x^7)^4+2} \right)\;×\dfrac {d}{dx}(x^7)\rightarrow\) Chain rule

\(=\dfrac {x^{14}+1}{x^{28}+1}×7x^6\)

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 

A

\(\dfrac {x^{14}+1}{x^{28}+1}×7x^6\)

.

B

\((x^2+1)×85x^2\)

C

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

D

\(5\,sin\,x\)

Option A is Correct

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 Questions

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

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

×

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

\(h(x)=x^2\) and \(\phi(x)=2-3x\)

\(\Rightarrow\) \(g'(x)=f(x^2)×\dfrac {d}{dx}(x^2)-f(2-3x)×\dfrac {d}{dx}(2-3x)\)

\(=(x^2\,cos\,x^2)×2x-(2-3x)cos(2-3x)×(-3)\ \)

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

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

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

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

Option A is Correct

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 Questions

Find the value of the integral \(\int\limits_{-3}^1\,(2x^2)\;dx\).

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

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

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

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

×

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

\(=2\int\limits_{-3}^1\,x^2\;dx \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left ( \int\,x^2\;dx=\dfrac {x^3}{3}+C\right)\)

\(=2×\dfrac {x^3}{3}\Bigg|_{-3}^1\)

\(=\dfrac {2}{3}[1^3-(-3)^3]\)

\(=\dfrac {2}{3}×28=\dfrac {56}{3}\)

Find the value of the integral \(\int\limits_{-3}^1\,(2x^2)\;dx\).

A

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

.

B

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

C

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

D

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

Option B is Correct

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 Questions

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

A 288

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

C –132

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

×

\(I=\int\limits_{4}^9\,\dfrac {t^2+1}{\sqrt t}\)

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

 

 

 

 

 

 

\(I= \int\limits_{4}^9\, t^{3/2}\;dt+ \int\limits_{4}^9\, t^{-1/2}\;dt\)

\(I= \dfrac {t^{5/2}}{5/2}\Bigg]_4^9+ \dfrac {t^{1/2}}{1/2}\Bigg]_4^9\)

\(I= \dfrac {2}{5}\Bigg[9^{5/2}-4^{5/2}\Bigg] +2\Bigg[9^{1/2}-4^{1/2}\Bigg]\)

\(I= \dfrac {2}{5} [243+32]+2[3-2]\)

\(I=\dfrac {422}{5}+2\)

\(I=\dfrac {432}{5}\)

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

A

288

.

B

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

C

–132

D

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

Option D is Correct

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