Q. 1 - Calculus

Question

Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample. 

(a) True or False: The absolute area between the graph of f and the x-axis on [a, b] is equal to$\left|{\int }_a^{b}f\right(x)dx|$.

(b) True or False: The area of the region between f(x) = x − 4 and g(x) = $-x^2$ on the interval [−3, 3] is negative.

(c) True or False: The signed area between the graph of f on [a, b] is always less than or equal to the absolute area on the same interval. 

(d) True or False: The area between any two graphs f and g on an interval [a, b] is given by ${\int }_a^b\left(f\right(x)-g(x\left)\right)dx$.

(e) True or False: The average value of the function f(x) = $x^2-3$ on [2, 6] is

$\frac{f\left(6\right)+f\left(2\right)}{2}$ = $\frac{33+1}{2}$ = 17.

(f) True or False: The average value of the function f(x) = $x^2-3$ on [2, 6] is $\frac{f\left(6\right)-f\left(2\right)}{4}$ = $\frac{33-1}{4}$= 8.

(g) True or False: The average value of f on [1, 5] is equal to the average of the average value of f on [1, 2] and the average value of f on [2, 5]. 

(h) True or False: The average value of f on [1, 5] is equal to the average of the average value of f on [1, 3] and the average value of f on [3, 5]. 

Step-by-Step Solution

Answer

Part (a). false

Part (b). false

Part (c). true

Part (d). false

Part (e). false

Part (f). false

Part (g). false

Part (h). true

Step 1. (a). Explanation

 $\text{ The statement is that the absolute area between the graph of }f\text{ and the }x\text{-axis on }[a,b]\text{ is }$

equal to $\left|{\int }_a^{b}f\right(x)dx|$.

$\text{ The absolute area between the graph of }f\text{ and the }x\text{-axis on }[a,b]\text{ is equal to }$${\int }_a^b\left|f\right(x\left)\right|dx$.

$\text{ Therefore, the statement is false. }$

Step 2. (b). Explanation

$\text{ The statement is the area of the region between }f\left(x\right)=x-4\text{ and }g\left(x\right)=-x^2\text{ on the interval }$

$[-3,3]\text{ is negative. }$

  The statement is false because the area is positive. 

  Therefore, the statement is false.

Step 3. (c). Explanation

$\text{ The statement is that the signed area between any two graphs of }f\text{ on }[a,b]\text{ is always less }$

than or equal to the absolute area on the same interval.

While calculating the absolute are, negative area is made positive so obviously the signed area$\text{ between any two graphs of }f\text{ on }[a,b]\text{ is always less than or equal to the absolute area on the }$

same interval.

Therefore, the statement is true.

Step 4. (d). Explanation

$\text{ The statement is that the area between any two graphs }f\text{ and }g\text{ on an interval }[a,b]\text{ is given }$

$\text{ by }{\int }_a^b \left(f\right(x)-g(x\left)\right)dx$.

$\text{ The statement is false. Because }\left(f\right(x)-g(x\left)\right)\text{ or }\left(g\right(x)-f(x\left)\right)\text{ depends on the graphs on }$

   that interval which has higher value than the other in that interval. 

   Therefore, the statement is false

Step 5. (e). Explanation

$\text{ The statement is that the average value of the function }f\left(x\right)=x^2-3\text{ on }[2,6]\text{ is }$

$\frac{f\left(6\right)+f\left(2\right)}{2}=\frac{33+1}{2}=17.$

$\text{ The statement is false because the exact average value of a function is }\frac{1}{b-a}{\int }_a^b f\left(x\right)dx\text{. }$

$\text{ So, the average value of the function }f\left(x\right)=x^2-3\text{ on }[2,6]\text{ is }$

$\frac{1}{6-2}{\int }_2^6 f\left(x\right)dx=\frac{1}{4}\left(F\right(6)-F(2\left)\right)$.

Therefore, the statement is false.

Step 6. (f). Explanation

$\text{ The statement is that the average value of the function }f\left(x\right)=x^2-3\text{ on }[2,6]\text{ is }$

$\frac{f\left(6\right)-f\left(2\right)}{4}=\frac{33-1}{4}=8$.

$\text{ The statement is false because the exact average value of a function is }\frac{1}{b-a}{\int }_a^b f\left(x\right)dx\text{. }$

$\text{ So, the average value of the function }f\left(x\right)=x^2-3\text{ on }[2,6]\text{ is }$

$\frac{1}{6-2}{\int }_2^6 f\left(x\right)dx=\frac{1}{4}\left(F\right(6)-F(2\left)\right)$.

Therefore, the statement is false.

Step 7. (g). Explanation

$\text{ The statement is that the average value of }f\text{ on }[1,5]\text{ is equal to the average of the average }$

$\text{ value of }f\text{ on }[1,2]\text{ and the average value of }f\text{ on }[2,5]\text{. }$

   The statement is false because 2 is not the average of 1,5 .

   Therefore, the statement is false.

Step 8. (h). Explanation

$\text{ The statement is that the average value or }f\text{ on }[1,5]\text{ is equal to the average of the average }$

$\text{ value of }f\text{ on }[1,3]\text{ and the average value of }f\text{ on }[3,5]\text{. }$

   The statement is true because 3 is the average of 1,5 .

   Therefore, the statement is true