Informative line

Application Of Ftc On Functions

Learn definite trigonometric integral of piecewise functions. Practice to Finding the area by using FTC 2, concavity of a curve.

Trigonometric Integrals

Basic Trigonometric anti derivative rules:

(1) $$\int sin\,x\;dx=-cos\,x+C$$                   (4)  $$\int cosec^2\,x\;dx=-cot\,x+C$$

(2) $$\int cos\,x\;dx=sin\,x+C$$                      (5) $$\int sec\,x\;tan\,x\;dx=sec\,x+C$$

(3) $$\int sec^2\,x\;dx=tan\,x+C$$                    (6) $$\int cosec\,x\;cot\,x\;dx=-cosec\,x+C$$

The value of  $$I=\int\limits_{\pi/3}^{\pi/2}\;sin\,x\;dx$$ is

A $$\dfrac {1}{2}$$

B $$\dfrac {3}{4}$$

C $$\dfrac {-1}{7}$$

D $$\dfrac {2}{3}$$

×

$$I=\int\limits_{\pi/3}^{\pi/2}\;sin\,x\;dx=-cos\,x\Bigg]_{\pi/3}^{\pi/2}$$

$$=-\left [cos\dfrac {\pi}{2}-cos\dfrac {\pi}{3} \right]$$

$$=-\left [0-\dfrac {1}{2} \right]=+\dfrac {1}{2}$$

$$\therefore \;I=\dfrac {1}{2}$$

The value of  $$I=\int\limits_{\pi/3}^{\pi/2}\;sin\,x\;dx$$ is

A

$$\dfrac {1}{2}$$

.

B

$$\dfrac {3}{4}$$

C

$$\dfrac {-1}{7}$$

D

$$\dfrac {2}{3}$$

Option A is Correct

Definite Integral of Piecewise Defined Functions

To find the definite integral of a piecewise function we split the integral into two integrand.

If $$f(x)=\begin{cases} g(x) & if & x<\alpha\\ h(x) & if &x \geq \alpha \end {cases}$$

then $$\int\limits_a^bf(x)\;dx= \int\limits_a^{\alpha} g(x)\;dx+ \int\limits_{\alpha}^{b} h(x)\;dx$$, where $$\alpha \in(a, b)$$

If  $$f(x)=\begin{cases} \sqrt x & if & 0\leq x\leq1\\ x & if &1< x\leq2 \end {cases}$$ then the value of $$I=\int\limits_0^2\,f(x)\;dx$$ is

A $$\dfrac {28}{3}$$

B $$\dfrac {13}{6}$$

C $$\dfrac {-7}{2}$$

D $$\dfrac {-8}{3}$$

×

$$I=\int\limits_0^2f(x)\;dx= \int\limits_0^{1} f(x)\;dx+ \int\limits_{1}^{2} f(x)\;dx=I_1+I_2$$

$$1$$ is the point at which definition of expression of f changes.

$$I_1= \int\limits_0^{1} f(x)\;dx$$

$$= \int\limits_{0}^{1} \sqrt x\;dx$$

$$=\dfrac {x^{\dfrac{3}{2}}}{\dfrac{3}{2}}\Bigg]_0^1$$

$$=\dfrac{2}{3}$$

$$I_2= \int\limits_1^{2} f(x)\;dx$$

$$= \int\limits_{1}^{2} x\;dx$$

$$=\dfrac {x^{2}}{2}\Bigg]_1^2$$

$$=\dfrac {4}{2}-\dfrac {1}{2}$$

$$=\dfrac {3}{2}$$

$$I=I_1+I_2$$

$$= \dfrac {2}{3}+\dfrac {3}{2}$$

$$=\dfrac {4+9}{6}$$

$$=\dfrac {13}{6}$$

If  $$f(x)=\begin{cases} \sqrt x & if & 0\leq x\leq1\\ x & if &1< x\leq2 \end {cases}$$ then the value of $$I=\int\limits_0^2\,f(x)\;dx$$ is

A

$$\dfrac {28}{3}$$

.

B

$$\dfrac {13}{6}$$

C

$$\dfrac {-7}{2}$$

D

$$\dfrac {-8}{3}$$

Option B is Correct

Evaluating Limits by making them Equivalent Definite Integrals

$$\int\limits_a^b\,f(x)\;dx=\lim\limits_{n\rightarrow \infty} \left ( \sum\limits_{i= 1}^n\;f(a+i\Delta x)(\Delta x) \right)$$ where $$\Delta x=\dfrac {b-a}{n}$$

• Now choose $$b=1,\;a=0$$ we get, $$\Delta x=\dfrac {1}{n}$$

$$\displaystyle\int\limits _0^1\,f(x)\;dx=\lim\limits_{n\rightarrow\infty}\;\left (\Sigma\,f\left (\dfrac {i}{n}\right)×\dfrac {1}{n}\right)$$

$$\therefore \;\lim\limits_{n\rightarrow\infty}\; \left (\sum\limits_{i=1}^{n}\dfrac {1}{n}\, f\left (\dfrac {i}{n}\right)\right)=\int\limits_0^1\,f(x)\;dx$$

• If we are given the limit we guess the function $$'f'$$ and evaluate the equivalent integral.

Evaluate the limit by recognizing it as an integral $$\lim\limits_{n\rightarrow\infty}\;\sum\limits_{i=1}^{n}\, \left (\dfrac {i^4}{n^5}\right)$$.

A $$\dfrac {2}{7}$$

B $$\dfrac {1}{5}$$

C $$\dfrac {18}{19}$$

D $$\dfrac {-1}{2}$$

×

Compare with standard form

$$\lim\limits_{n\rightarrow\infty}\;\sum\limits_{i=1}^{n}\, \left (\dfrac {i^4}{n^5}\right)= \lim\limits_{n\rightarrow\infty}\;\sum\limits_{i=1}^{n}\,\dfrac {1}{n} f\left (\dfrac {i}{n}\right)$$

$$\lim\limits_{n\rightarrow\infty}\;\sum\limits_{i=1}^{n}\,\dfrac {1}{n}\, \left (\dfrac {i}{n}\right)^4= \lim\limits_{n\rightarrow\infty}\;\sum\limits\,\dfrac {1}{n} f\left (\dfrac {i}{n}\right)$$ (take $$\dfrac {1}{n}$$ separately to get the form)

$$f\left ( \dfrac {i}{n} \right)= \left ( \dfrac {i}{n} \right)^4\Rightarrow f(x)=x^4$$

$$\therefore \;\lim\limits_{n\rightarrow\infty}\; \left (\sum\limits_{i=1}^{n}\dfrac {1}{n}\, \left (\dfrac {i}{n}\right)^4\right)=\int\limits_0^1\,x^4\;dx$$

$$=\Bigg[\dfrac {x^5}{5}\Bigg]_0^1=\dfrac {1}{5}$$

Evaluate the limit by recognizing it as an integral $$\lim\limits_{n\rightarrow\infty}\;\sum\limits_{i=1}^{n}\, \left (\dfrac {i^4}{n^5}\right)$$.

A

$$\dfrac {2}{7}$$

.

B

$$\dfrac {1}{5}$$

C

$$\dfrac {18}{19}$$

D

$$\dfrac {-1}{2}$$

Option B is Correct

Integrand Functions which are Discontinuous in the Interval with Lower and Upper limits as End Points

If f is discontinuous in [a, b] then $$\int\limits_a^b\;f(x)\;dx=F(b)-F(a)$$ cannot be applied.

Consider

$$\displaystyle\int_{-1}^2\dfrac {1}{x^2}\;dx=-\dfrac {1}{x}\Bigg]_{-1}^2=-\Bigg[\dfrac {1}{2}-1\Bigg]=\dfrac {1}{2}$$

• The above calculation is wrong because $$\dfrac {1}{x^2}$$ is a positive function and the value of integral cannot be negative.
• This is because $$\dfrac {1}{x^2}$$ is discontinuous at $$x=0$$ and  here in [–1, 2 ].

What is wrong with the following result? $$\displaystyle\int\limits_{-1}^2\dfrac {1}{x^4}\;dx$$ $$=\dfrac {-1}{3}x^{-3}\Bigg]_{-1}^{2}$$ $$=\dfrac {-1}{3}\Big[2^{-3}-(-1)^{-3}\Big]$$ $$=\dfrac {-1}{3}\Big[\dfrac {1}{8}+1\Big]$$ $$=\dfrac {-1}{3}×\dfrac {9}{8}=\dfrac {-3}{8}$$ $$\therefore\;\displaystyle\int\limits_{-1}^{2}\dfrac {1}{x^4}\;dx=\dfrac {-3}{8}$$

A $$f(x)=\dfrac {1}{x^4}$$ is discontinuous at $$x=1$$ so $$f$$ is not integrable in [–1, 2]

B $$f(x)=\dfrac {1}{x^4}$$ is discontinuous at $$x=0$$ so $$f$$ is not integrable in [–1, 2]

C $$f(x)=\dfrac {1}{x^4}$$ is discontinuous at $$x=\dfrac {1}{2}$$ so $$f$$ is not integrable in [–1, 2]

D $$f(x)=\dfrac {1}{x^4}$$ is discontinuous at $$x=-\dfrac {1}{2}$$ so $$f$$ is not integrable in [–1, 2]

×

$$f(x)=\dfrac {1}{x^4}$$ has a point of discontinuity at  $$x=0$$.

Since,  $$0\in[-1, 2]$$

$$\therefore\;f(x)$$ is discontinuous  in [–1, 2] hence$$'f'$$ is not integrable.

Integral can't be negative as $$f(x)=\dfrac {1}{x^4}$$ is a positive function.

What is wrong with the following result? $$\displaystyle\int\limits_{-1}^2\dfrac {1}{x^4}\;dx$$ $$=\dfrac {-1}{3}x^{-3}\Bigg]_{-1}^{2}$$ $$=\dfrac {-1}{3}\Big[2^{-3}-(-1)^{-3}\Big]$$ $$=\dfrac {-1}{3}\Big[\dfrac {1}{8}+1\Big]$$ $$=\dfrac {-1}{3}×\dfrac {9}{8}=\dfrac {-3}{8}$$ $$\therefore\;\displaystyle\int\limits_{-1}^{2}\dfrac {1}{x^4}\;dx=\dfrac {-3}{8}$$

A

$$f(x)=\dfrac {1}{x^4}$$ is discontinuous at $$x=1$$ so $$f$$ is not integrable in [–1, 2]

.

B

$$f(x)=\dfrac {1}{x^4}$$ is discontinuous at $$x=0$$ so $$f$$ is not integrable in [–1, 2]

C

$$f(x)=\dfrac {1}{x^4}$$ is discontinuous at $$x=\dfrac {1}{2}$$ so $$f$$ is not integrable in [–1, 2]

D

$$f(x)=\dfrac {1}{x^4}$$ is discontinuous at $$x=-\dfrac {1}{2}$$ so $$f$$ is not integrable in [–1, 2]

Option B is Correct

Finding the area by using FTC 2

The area bounded by $$y=f(x)$$ and $$x$$-axis between $$x=a$$ and $$x=b$$ is given by

$$A=\int\limits_a^b\,f(x)\;dx$$

• We have an easy formula in the form of $$=\int\limits_a^b\,f(x)\;dx=F(b)-F(a)$$, where $$'F'$$is the anti derivative of $$'f'$$.
• The area we find by the formulae above will be in square units where the unit will be the unit we choose on $$x$$ and $$y$$axis. Suppose the units on $$x$$ and $$y$$ axis are cm, then the area is in  $$cm^2$$ . We often ignore the units and express the area as a number.

Find the area bounded by $$y=\dfrac {1}{x^2}$$, $$x$$-axis between $$x=2$$ and $$x=5$$.

A $$\dfrac {2}{5}$$

B $$\dfrac {18}{71}$$

C $$\dfrac {-1}{5}$$

D $$\dfrac {3}{10}$$

×

$$A=\displaystyle\int\limits_2^5\dfrac {1}{x^2}\;dx$$

$$=\int\limits_2^5\,x^{-2}\;dx$$

$$=\dfrac {-1}{x}\Bigg]_2^5$$

$$=-\left [ \dfrac {1}{5}- \dfrac {1}{2} \right]$$

$$=\dfrac {3}{10}$$

Find the area bounded by $$y=\dfrac {1}{x^2}$$, $$x$$-axis between $$x=2$$ and $$x=5$$.

A

$$\dfrac {2}{5}$$

.

B

$$\dfrac {18}{71}$$

C

$$\dfrac {-1}{5}$$

D

$$\dfrac {3}{10}$$

Option D is Correct

Concavity of Function given in integral form (Concave Upward)

A curve is said to be concave upward if tangent at every point on it lies below the curve.

Two situations are shown in which the curve is concave upwards.

We say the curve $$y=f(x)$$ is concave upwards in [a, b] as tangent at each point lies below the curve.

• Concave upward  $$\Rightarrow \dfrac {d^2y}{dx^2}>0$$ or $$f''(x)>0$$
• If $$g(x)=\int\limits_a^{x}\,f(t)\;dt$$, then $$g''(x)>0\Rightarrow f'(x)>0$$

Given is the graph of a function $$'f'$$. Let $$g(x)=\int\limits_0^{x}\;f(t)\;dt$$. On what interval is $$'g'$$ concave upwards?

A $$(1, 7)$$

B $$(0, 2)\cup(3.5,6)$$

C $$(1, 5)$$

D $$(-2, 8)$$

×

Concave upwards $$\Rightarrow g''(x)>0$$

Given, $$g(x)=\int\limits_a^{x}\;f(x)\;dx$$, then $$g'(x)=f(x)$$ by FTC 1

$$g'(x) = f(x)$$ and $$g '' (x) = f'(x)$$

So, $$g'' (x) > 0$$ $$\Rightarrow f'(x) > 0 \Rightarrow\, 'f'$$  is increasing.

$$\therefore \; x\in(0, 2)\cup(3.5,6)$$

Given is the graph of a function $$'f'$$. Let $$g(x)=\int\limits_0^{x}\;f(t)\;dt$$. On what interval is $$'g'$$ concave upwards?

A

$$(1, 7)$$

.

B

$$(0, 2)\cup(3.5,6)$$

C

$$(1, 5)$$

D

$$(-2, 8)$$

Option B is Correct

Concavity of a Function given in Integral Form (Concave Downward)

• A curve is said to be concave downward if tangent at every point of the curve lies above the curve.

Two situations are shown in which curve is concave downward.

We say the curve $$y=f(x)$$ is concave downwards in [a,b] as tangent lies above the curve.

• Concave downwards $$\Rightarrow \dfrac {d^2y}{dx^2}<0\Rightarrow f''(x)<0$$
• If $$g(x)=\int\limits_a^x\;f(t)\;dt$$, then $$g''(x)<0\;\Rightarrow f'(x)<0$$

Given is the graph of a function $$'f'$$. Let $$g(x)=\int\limits_0^{x}\;f(t)\;dt$$. On what interval is $$'g'$$ concave downwards?

A $$x\in(2, 3.5)\cup(6,8)$$

B $$x\in(1,3)\cup(5,6)$$

C $$x\in(1,7)$$

D $$x\in(2,4)\cup(6,8)$$

×

Concave downwards $$\Rightarrow g''(x)<0$$

$$g'(x)=f(x)$$ and $$g''(x)=f'(x)$$

$$g''(x)<0\;\Rightarrow f'(x)<0\Rightarrow f$$ is decreasing

$$\therefore\,x\in(2, 3.5)\cup(6,8)$$

Given is the graph of a function $$'f'$$. Let $$g(x)=\int\limits_0^{x}\;f(t)\;dt$$. On what interval is $$'g'$$ concave downwards?

A

$$x\in(2, 3.5)\cup(6,8)$$

.

B

$$x\in(1,3)\cup(5,6)$$

C

$$x\in(1,7)$$

D

$$x\in(2,4)\cup(6,8)$$

Option A is Correct

Concavity of a Curve

• If the graph of function  $$'f'$$  lies above all its tangents in an interval then it is called concave upwards on that interval.
• If $$f''(x)>0$$ for all  $$x$$ in an interval $$[a,b]$$ then we say that graph of $$f$$ is concave upwards on the interval.

• If the graph of function $$'f'$$ lies below all its tangents  in an  interval , it is called concave downwards in that interval.
• If $$f''(x)<0$$ for all $$x$$ in an interval $$[a,b]$$ then we say that graph of $$f$$ is concave downwards on the interval.

Find the intervals of concavity of $$f(x)=4x^3+3x^2-6x+1$$.

A Concave upwards on $$(-\infty,-1)$$ and downwards on $$(-1,\infty)$$

B Concave upwards on $$\Big(-\infty,\dfrac {2}{3}\Big)$$ and downwards on $$\Big(\dfrac {2}{3},\,\infty\Big)$$

C Concave upwards on $$\Big(\dfrac {-1}{4},\,\infty\Big)$$ and downwards on $$\Big(-\infty,\,\dfrac {-1}{4}\Big)$$

D Concave upwards on$$(-2,\infty)$$  and downwards on $$(-\infty,-2)$$

×

For concave upward graph$$\rightarrow f''>0$$

For concave downward graph $$\rightarrow f''<0$$

In this case,

$$f(x)=4x^3+3x^2-6x+1$$

$$\therefore\,\,f'(x)=12x^2+6x-6$$ and  $$\,f''(x)=24x+6$$

$$\therefore\,\,f''(x)>0$$

$$\Rightarrow 24x+6>0$$

$$\Rightarrow x>\dfrac {-1}{4}$$

$$\Rightarrow x\in \Big(\dfrac {-1}{4},\infty\Big)$$

$$\therefore$$ Concave upwards on $$x\in \Big(\dfrac {-1}{4},\infty\Big)$$

and concave downward on $$x\in \Big(-\infty,\,\dfrac {-1}{4}\Big)$$

Find the intervals of concavity of $$f(x)=4x^3+3x^2-6x+1$$.

A

Concave upwards on $$(-\infty,-1)$$ and downwards on $$(-1,\infty)$$

.

B

Concave upwards on $$\Big(-\infty,\dfrac {2}{3}\Big)$$ and downwards on $$\Big(\dfrac {2}{3},\,\infty\Big)$$

C

Concave upwards on $$\Big(\dfrac {-1}{4},\,\infty\Big)$$ and downwards on $$\Big(-\infty,\,\dfrac {-1}{4}\Big)$$

D

Concave upwards on$$(-2,\infty)$$  and downwards on $$(-\infty,-2)$$

Option C is Correct