Question 1 |
Consider the definite integral
\int_{1}^{2}(4x^2+2x+6)dx
Let I_e be the exact value of the integral. If the same integral is estimated using Simpson's rule with 10 equal subintervals, the value is I_s . The percentage error is defined as e=100 \times (I_e-I_s)/I_e . The value of e is
\int_{1}^{2}(4x^2+2x+6)dx
Let I_e be the exact value of the integral. If the same integral is estimated using Simpson's rule with 10 equal subintervals, the value is I_s . The percentage error is defined as e=100 \times (I_e-I_s)/I_e . The value of e is
2.5 | |
3.5 | |
1.2 | |
0 |
Question 1 Explanation:
Exact value
\begin{aligned} &=\int_{1}^{2} (4x^2+2x+6)dx\\ &=\frac{4x^3}{3}+\frac{2x^2}{2}+6x\\ &=\frac{4}{3}(\pi) \times (3)+6\\ &=\frac{28}{3}+9=\frac{55}{3} \end{aligned}
Approximate value
Here f(x) is a polynomial of degree 2, so Simpsons rule gives exact value with zero error
\begin{aligned} \therefore \;\; \text{Approx value}&=\frac{55}{3}\\ \frac{I_e-I_s}{I_e}&=0\\ \therefore \;\;e=\left (\frac{I_e-I_s}{I_e} \right )\times 100&=0 \end{aligned}
\begin{aligned} &=\int_{1}^{2} (4x^2+2x+6)dx\\ &=\frac{4x^3}{3}+\frac{2x^2}{2}+6x\\ &=\frac{4}{3}(\pi) \times (3)+6\\ &=\frac{28}{3}+9=\frac{55}{3} \end{aligned}
Approximate value
Here f(x) is a polynomial of degree 2, so Simpsons rule gives exact value with zero error
\begin{aligned} \therefore \;\; \text{Approx value}&=\frac{55}{3}\\ \frac{I_e-I_s}{I_e}&=0\\ \therefore \;\;e=\left (\frac{I_e-I_s}{I_e} \right )\times 100&=0 \end{aligned}
Question 2 |
Value of \int_{4}^{5.2}\ln x \; dx using Simpson's one-third rule with interval size 0.3 is
1.83 | |
1.6 | |
1.51 | |
1.06 |
Question 2 Explanation:
\begin{aligned} \text { Here } \qquad f(x)&=\log x\\ a=4, \quad b=5.2, h=0.3\\ \text { So, } \qquad n&=\frac{b-a}{h}=\frac{5.2-4}{0.3}=4\\ \end{aligned}
\begin{array}{cccccc} \hline x & 4 & 4.3 & 4.6 & 4.9 & 5.2 \\ \hline y & \log 4 & \log 4.3 & \log 4.6 & \log 4.9 & \log 5.2 \\ \hline & y_{0} & y_{1} & y_{2} & y_{3} & y_{4} \end{array}
As per Simpson's 1 / 3^{\text {rd }} rule
\begin{aligned} \int_{4}^{5.2} \log x d x &=\frac{h}{3}\left[y_{0}+y_{4}+4\left(y_{1}+y_{3}\right)+2\left(y_{2}\right)\right] \\ &=1.8272 \simeq 1.83 \end{aligned}
\begin{array}{cccccc} \hline x & 4 & 4.3 & 4.6 & 4.9 & 5.2 \\ \hline y & \log 4 & \log 4.3 & \log 4.6 & \log 4.9 & \log 5.2 \\ \hline & y_{0} & y_{1} & y_{2} & y_{3} & y_{4} \end{array}
As per Simpson's 1 / 3^{\text {rd }} rule
\begin{aligned} \int_{4}^{5.2} \log x d x &=\frac{h}{3}\left[y_{0}+y_{4}+4\left(y_{1}+y_{3}\right)+2\left(y_{2}\right)\right] \\ &=1.8272 \simeq 1.83 \end{aligned}
Question 3 |
For the integral \int_{0}^{\pi/2}(8+4 \cos x)dx, the absolute percentage error in numerical evaluation
with the Trapezoidal rule, using only the end points, is ________. (round off to one decimal
place).
15.7 | |
85.9 | |
0.859 | |
5.2 |
Question 3 Explanation:
True value
\begin{aligned} \int_{0}^{\pi / 2}(8+4 \cos x) d x &=[8 x+(4 \sin x)]_{0}^{\pi / 2} \\ &=4 \pi+4=16.566 \end{aligned}
By trapezoidal rule, (single step)
\begin{aligned} h&=\frac{\pi}{2} \\ \text{Approx.} \qquad I&=\frac{\pi}{4}[12+8]=5 \pi=15.707 \end{aligned}
Absolute error = |True value - Approximate value|
\begin{aligned} &=|16.566-15.707|=0.859 \\ \text { Absolute percentage error } &=\frac{0.859}{16.566} \times 100=5.18 \% \approx 5.2 \% \end{aligned}
\begin{aligned} \int_{0}^{\pi / 2}(8+4 \cos x) d x &=[8 x+(4 \sin x)]_{0}^{\pi / 2} \\ &=4 \pi+4=16.566 \end{aligned}
By trapezoidal rule, (single step)
\begin{aligned} h&=\frac{\pi}{2} \\ \text{Approx.} \qquad I&=\frac{\pi}{4}[12+8]=5 \pi=15.707 \end{aligned}
Absolute error = |True value - Approximate value|
\begin{aligned} &=|16.566-15.707|=0.859 \\ \text { Absolute percentage error } &=\frac{0.859}{16.566} \times 100=5.18 \% \approx 5.2 \% \end{aligned}
Question 4 |
The evaluation of the definite integral \int_{-1}^{1.4}x|x|dx by using Simpson's 1/3^{rd} (one-third) rule
with step size h = 0.6 yields
0.914 | |
1.248 | |
0.581 | |
0.592 |
Question 4 Explanation:
\int_{-1}^{1.4} x|x| d x \text { step size } h=0.6
Simpson's 1 / 3^{\text {rd }} rule
\begin{aligned} \int_{-1}^{1.4} x|x| d x &=\frac{0.6}{3}[(-1+1.96)+4(-0.16+0.64)+2(0.04)] \\ &=0.2[0.96+1.92+0.08]=0.592 \end{aligned}
Simpson's 1 / 3^{\text {rd }} rule
\begin{aligned} \int_{-1}^{1.4} x|x| d x &=\frac{0.6}{3}[(-1+1.96)+4(-0.16+0.64)+2(0.04)] \\ &=0.2[0.96+1.92+0.08]=0.592 \end{aligned}
Question 5 |
Evaluation of \int_{2}^{4}x^3 dx using a 2-equal-segment trapezoidal rule gives a value of______
32 | |
88 | |
63 | |
42 |
Question 5 Explanation:
\begin{aligned} &\text { Let } y=f(x)=x^{3}\\ &\begin{array}{|c|c|c|c|} \hline \mathrm{x} & 2 & 3 & 4 \\ \hline \mathrm{y} & 8 & 27 & 64 \\ \hline \end{array} \end{aligned}
Given n=2 \Rightarrow h=\frac{4-2}{2}=1
By trapezoidal rule, we have
\int_{2}^{4} x^{3} d x=\frac{h}{2}\left\{\left(y_{0}+y_{n}\right)+2\left(y_{1}+y_{2}+\ldots .+y_{n-1}\right)\right\}=\frac{1}{2}\{(8+64)+2(27)\}=63
Given n=2 \Rightarrow h=\frac{4-2}{2}=1
By trapezoidal rule, we have
\int_{2}^{4} x^{3} d x=\frac{h}{2}\left\{\left(y_{0}+y_{n}\right)+2\left(y_{1}+y_{2}+\ldots .+y_{n-1}\right)\right\}=\frac{1}{2}\{(8+64)+2(27)\}=63
There are 5 questions to complete.
Reconstruct the question number 27….