Question
Find the area under the curve of \(f(x)=(x-1)^3\) from \(x=0\) to \(x=2\).
(A) \(0\)
(B) \(\frac{1}{4}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{1}{16}\)
(E) None of these
▶️Answer/Explanation
Ans:C
Net area from x=0 to x=2 is equal to sum of $A_1$ and $A_2$
$A_{\rm{net}}=A_1+A_2$
\( \begin{aligned} & A_1=\int_0^1(x-1)^3 d x=\left.\frac{(x-1)^4}{4}\right|_0 ^1=0-\frac{1}{4} \\ & A_1=-\frac{1}{4} \Rightarrow A_1=\frac{1}{4}\quad\left[A \geqslant0\right]\end{aligned} \)
Similarly, \( \begin{aligned} & A_2=\int_1^2(x-1)^3 d x=\left.\frac{(x-1)^4}{4}\right|_1 ^2=\frac{1}{4}-0 \\ & A_2=\frac{1}{4}\end{aligned} \)
$A_{\rm{net}}=\frac{1}{4}+\frac{1}{4}\Rightarrow \frac{1}{2}$
Question
The area of the shaded region in the figure above is represented by which of the following integrals?
(A)\(\int_{a}^{c}\left ( |f(x)|-g(x) |\right )dx\)
(B)\(\int_{b}^{c} f(x)dx-\int_{a}^{c}g(x)dx\)
(C)\(\int_{a}^{c}(g(x)-f(x))dx\)
(D)\(\int_{a}^{c}(f(x)-g(x))dx\)
(E)\(\int_{a}^{b}(g(x)-f(x))dx+\int_{b}^{c}(f(x)-g(x))dx\)
▶️Answer/Explanation
Ans:D
The interval is x = a to x = c. The height of a rectangular slice is the top curve, f(x), minus the bottom curve, g(x). The area of the rectangular slice is therefore \((f(x) − g(x))\Delta x\). Set up a Riemann sum and take the limit as \(\Delta x\) goes to 0 to get a definite integral.
Question
Which of the following represents the area of the shaded region in the figure above?
(A) \(\int_{c}^{d}f(y)dy\) (B)\(\int_{a}^{b}(d-f(x))dx \) (C)\(f'(b)-f'(a) \) (D)\( (b-a)[f(b)-f(a)] \) (E)\((d-c)[f(b)-f(a)]\)
▶️Answer/Explanation
Ans:B
Summing pieces of the form: (vertical). (small width), vertical = (d− f(x)), width = \(\Delta x\)
Area=\(\int_{a}^{b}(d-f(x))dx\)
Question
The function f is continuous on the closed interval [ 0, 6] and has the values given in the table above. The trapezoidal approximation for\( \int_{0}^{6}f(x)\) dx found with 3 subintervals of equal length is 52. What is the value of k ?
(A) 2 (B) 6 (C) 7 (D) 10 (E) 14
▶️Answer/Explanation
Ans:D