Informative line

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

(Put $$x=2$$) $$= \underbrace {\left(\dfrac {1}{2}×(2+1)×1\right)}_{\text {trapezium OABC}}+ \underbrace {\left(\dfrac {1}{2}×(1+3)×1\right)}_{\text {trapezium ABED}}$$ $$=\dfrac {1}{2}×3 + \dfrac {1}{2}×4$$

$$=\dfrac {7}{2}$$ ### 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

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.

#### 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 (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.

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

#### 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

# 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

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

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

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

#### 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

# 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 $$x$$.

#### 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

# 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 $$C=0$$.

#### 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