Question 1 |
A function f(x), that is smooth and convex-shaped between interval \left(x_{1}, x_{u}\right) is shown in the figure. This function is observed at odd number of regularly spaced points. If the area under the function is computed numerically, then ____
the numerical value of the area obtained using
the trapezoidal rule will be less than the
actual. | |
the numerical value of the area obtained using
the trapezoidal rule will be more than the
actual. | |
the numerical value of the area obtained using
the trapezoidal rule will be exactly equal to
the actual. | |
with the given details, the numerical value of
area cannot be obtained using trapezoidal rule |
Question 1 Explanation:
Approximated function has under estimation so numerical value of the area obtained using trapezoidal rule will be less than the actual.
Question 2 |
Consider the following recursive iteration scheme for different values of variable P with the initial guess x_1=1:
x_{n+1}=\frac{1}{2}\left ( x_n+\frac{P}{x_n} \right ),\;\;\;n=1,2,3,4,5
For P=2,x_5 is obtained to be 1.414, rounded-off to three decimal places. For P=3,x_5 is obtained to be 1.732, rounded-off to three decimal places.
If P=10, the numerical value of x_5 is _________ . (round off to three decimal places)
x_{n+1}=\frac{1}{2}\left ( x_n+\frac{P}{x_n} \right ),\;\;\;n=1,2,3,4,5
For P=2,x_5 is obtained to be 1.414, rounded-off to three decimal places. For P=3,x_5 is obtained to be 1.732, rounded-off to three decimal places.
If P=10, the numerical value of x_5 is _________ . (round off to three decimal places)
2.155 | |
3.162 | |
1.125 | |
4.568 |
Question 2 Explanation:
x_{n+1}=\frac{1}{2}\left ( x_n+\frac{P}{x_n} \right )
Converges when x_{n+1}=x_{n}=\alpha
\begin{aligned} \alpha &=\frac{1}{2}\left ( \alpha+\frac{P}{\alpha} \right )\\ \frac{\alpha}{2}&=\frac{P}{2\alpha}\\ \alpha&=\sqrt{P} \end{aligned}
When P=2, x_5=\sqrt{2}=1.4124
When P=3, x_5=\sqrt{3}=1.732
When P=10, x_5=\sqrt{10}=3.162
Converges when x_{n+1}=x_{n}=\alpha
\begin{aligned} \alpha &=\frac{1}{2}\left ( \alpha+\frac{P}{\alpha} \right )\\ \frac{\alpha}{2}&=\frac{P}{2\alpha}\\ \alpha&=\sqrt{P} \end{aligned}
When P=2, x_5=\sqrt{2}=1.4124
When P=3, x_5=\sqrt{3}=1.732
When P=10, x_5=\sqrt{10}=3.162
Question 3 |
Numerically integrate, f(x)=10 x-20 x^{2} from lower limit a=0 to upper limit b=0.5. Use Trapezoidal rule with five equal subdivisions. The value (in units,round off to two decimal places) obtained is ____________
0.78 | |
0.65 | |
0.4 | |
0.56 |
Question 3 Explanation:
\begin{aligned} y&=10 x-20 x^{2} \\ a&=0, b=0.5, n=5 \\ \text { So, } \qquad \qquad \qquad h&=\frac{b-a}{n}=0.1 \end{aligned}
And
\begin{aligned} I &=\int_{0}^{0.5} f(x) d x=\frac{h}{2}\left[y_{0}+y_{5}+2\left(y_{1}+y_{2}+y_{3}+y_{4}\right)\right] \\ &=\frac{0.1}{2}[0+0+2(0.8+1.2+1.2+0.8)] \\ &=0.40 \end{aligned}
And
\begin{aligned} I &=\int_{0}^{0.5} f(x) d x=\frac{h}{2}\left[y_{0}+y_{5}+2\left(y_{1}+y_{2}+y_{3}+y_{4}\right)\right] \\ &=\frac{0.1}{2}[0+0+2(0.8+1.2+1.2+0.8)] \\ &=0.40 \end{aligned}
Question 4 |
The value of \int_{0}^{1} e^{x} d x using the trapezoidal rule with four equal subintervals is
1.718 | |
1.727 | |
2.192 | |
2.718 |
Question 4 Explanation:
Let \int_{a}^{b}f(x)dx=\int_{0}^{1}e^x \; dx \text{ and }n=4
Then a=0,b=1,f(x)=e^x \text{ and }h=\frac{b-a}{n}=\frac{1-0}{4}=0.25
\begin{array}{|c|c|c|c|c|c|} \hline x&0&0.25&0.50&0.75&1\\ \hline y=f(x=e^x)&1&1.284&1.649&2.117&2.718\\ \hline \end{array}
The formula of trapezoidal rule to the given data is given by
\int_{a}^{b}f(x)dx\simeq \int_{a}^{b}p(x)dx=\frac{h}{2}\left \{ (y_0+y_n)+2(y_1+y_2+y_3) \right \}
\int_{0}^{1}e^x \; dx\simeq \int_{0}^{1}p(x)dx=\frac{0.25}{2}\left \{ (1+2.718)+2(1.284+1.640+2.117) \right \}
\int_{0}^{1}e^x \; dx\simeq \int_{0}^{1}p(x)dx=\frac{0.25}{2}\left \{ (3.718)+2(5.03) \right \}=1.727
Then a=0,b=1,f(x)=e^x \text{ and }h=\frac{b-a}{n}=\frac{1-0}{4}=0.25
\begin{array}{|c|c|c|c|c|c|} \hline x&0&0.25&0.50&0.75&1\\ \hline y=f(x=e^x)&1&1.284&1.649&2.117&2.718\\ \hline \end{array}
The formula of trapezoidal rule to the given data is given by
\int_{a}^{b}f(x)dx\simeq \int_{a}^{b}p(x)dx=\frac{h}{2}\left \{ (y_0+y_n)+2(y_1+y_2+y_3) \right \}
\int_{0}^{1}e^x \; dx\simeq \int_{0}^{1}p(x)dx=\frac{0.25}{2}\left \{ (1+2.718)+2(1.284+1.640+2.117) \right \}
\int_{0}^{1}e^x \; dx\simeq \int_{0}^{1}p(x)dx=\frac{0.25}{2}\left \{ (3.718)+2(5.03) \right \}=1.727
Question 5 |
The integral
\int_{0}^{1}(5x^3+4x^2+3x+2)dx
is estimated numerically using three alternative methods namely the rectangular, trapezoidal and Simpson's rules with a common step size. In this context, which one of the following statement is TRUE?
\int_{0}^{1}(5x^3+4x^2+3x+2)dx
is estimated numerically using three alternative methods namely the rectangular, trapezoidal and Simpson's rules with a common step size. In this context, which one of the following statement is TRUE?
Simpsons rule as well as rectangular rule of estimation will give NON-zero error. | |
Simpson's rule, rectangular rule as well as trapezoidal rule of estimation will give
NON-zero error. | |
Only the rectangular rule of estimation will given zero error. | |
Only Simpson's rule of estimation will give zero error. |
Question 5 Explanation:
Because integral is a polynomial of 3rd degree so Simpson's rule will give error free answer.
There are 5 questions to complete.