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
Question 6 |
An explicit forward Euler method is used to numerically integrate the differential equation
\frac{\mathrm{d} y}{\mathrm{d} x}=y
using a time step of 0.1. With the initial condition y(0) = 1, the value of y(1) computed by this method is ___________ (correct to two decimal places).
\frac{\mathrm{d} y}{\mathrm{d} x}=y
using a time step of 0.1. With the initial condition y(0) = 1, the value of y(1) computed by this method is ___________ (correct to two decimal places).
2.59 | |
1.54 | |
2.24 | |
3.67 |
Question 6 Explanation:
\begin{aligned} y_{1} &=y_{0}+h_{A}\left(t_{0}, y_{0}\right) \\ &=y_{0}+h y_{0} \\ &=1+0.1(1) \\ y_{1} &=1.1 \\ y_{2} &=y_{1}+h_{f}\left(t_{1}, y_{1}\right) \\ &=y_{1}+h . y_{1} \\ &=1.1+0.1(1.1) \\ y_{2} &=1.21 \\ y_{3} &=y_{2}+h f\left(t_{2,} y_{2}\right) \\ &=y_{2}+h . y_{2} \\ &=1.21+0.1 \times 1.21 \\ y_{3} &=1.331 \\ y_{4} &=y_{3}+h . f\left(t_{3}, y_{3}\right) \\ &=y_{3}+h . y_{3} \\ &=1.331+0.1 \times 1.331 \\ y_{4} &=1.4641 \\ y_{5} &=y_{4}+h . f\left(t_{4}, y_{4}\right) \\ &=y_{4}+h . y_{4} \\ &=1.4641+0.1 \times(1.4641) \\ &=1.61051 \\ y_{6} &=y_{5}+h . f\left(t_{5}, y_{5}\right) \\ &=y_{5}+h . y_{5} \\ &=1.61051+0.1 \times 1.61051 \end{aligned}
\begin{aligned} y_{6} &=1.771561 \\ y_{7} &=y_{6}+h . f\left(t_{6}, y_{6}\right) \\ &=y_{6}+h \times y_{6} \\ &=1.771561+0.1 \times 1.771561=1.9487 \\ y_{8} &=y_{7}+h . f\left(t_{7}, y_{7}\right) \\ &=y_{7}+h . y_{7} \\ &=1.9487+0.1 \times(1.9487) \\ y_{8} &=2.14357 \\ y_{9} &=y_{8}+h . f\left(t_{8}, y_{8}\right) \\ &=y_{8}+h . y_{8}=2.14357+0.1 \times 2.14357 \\ y_{9} &=2.3579 \\ y_{10} &=y_{9}+h . f\left(t_{9}, y_{9}\right)=y_{9}+h . y_{9} \\ &=2.3579+0.1 \times(2.3579) \\ y_{10} &=2.5937 \end{aligned}
\begin{aligned} y_{6} &=1.771561 \\ y_{7} &=y_{6}+h . f\left(t_{6}, y_{6}\right) \\ &=y_{6}+h \times y_{6} \\ &=1.771561+0.1 \times 1.771561=1.9487 \\ y_{8} &=y_{7}+h . f\left(t_{7}, y_{7}\right) \\ &=y_{7}+h . y_{7} \\ &=1.9487+0.1 \times(1.9487) \\ y_{8} &=2.14357 \\ y_{9} &=y_{8}+h . f\left(t_{8}, y_{8}\right) \\ &=y_{8}+h . y_{8}=2.14357+0.1 \times 2.14357 \\ y_{9} &=2.3579 \\ y_{10} &=y_{9}+h . f\left(t_{9}, y_{9}\right)=y_{9}+h . y_{9} \\ &=2.3579+0.1 \times(2.3579) \\ y_{10} &=2.5937 \end{aligned}
Question 7 |
P(0,3), Q(0.5, 4), and R (1,5) are three points on the curve defined by f(x). Numerical integration is carried out using both Trapezoidal rule and Simpson's rule within limits x = 0 and x =1 for the curve. The difference between the two results will be.
0 | |
0.25 | |
0.5 | |
1 |
Question 7 Explanation:
By Trapezoidal rule
\begin{aligned} \int_{0}^{1} f(x) d x &=\frac{h}{2}\left[\left(y_{0}+y_{2}\right)+2 y_{1}\right] \\ &=\frac{0.5}{2}[(3+5)+2(4)]=4 \end{aligned}
By Simpson rule
\int_{0}^{1} f(x) d x=\frac{h}{3}\left[\left(y_{0}+y_{2}\right)+4 y_{1}\right]
=\frac{0.5}{3}[(3+5)+4(4)]=4
The difference between the two results will be zero.
Question 8 |
The root of the function f(x)=x^{3}+x-1 obtained after first iteration on application of NewtonRaphson scheme using an initial guess of x_{0}=1 is
0.682 | |
0.686 | |
0.75 | |
1 |
Question 8 Explanation:
\begin{aligned} f(x)&=x^3+x-1\\ f(1)&=1+1-1=1 \\ f'(x)&=3x^2+1\\ f'(1)&=3+1=4\\ x_1&=x_0-\frac{f(x_0)}{f'(x_0)}\\ &=1-\frac{1}{4}=1-0.25=0.75 \end{aligned}
Question 9 |
The error in numerically computing the integral \int_{0}^{\pi }(\sin x+\cos x )dx using the trapezoidal rule with three intervals of equal length between 0 and \pi is
0.186 | |
0.12121 | |
0.9898 | |
0.6587 |
Question 9 Explanation:
\begin{array}{|c|c|c|c|c|} \hline x & 0 & \frac{\pi}{3} & \frac{2 \pi}{3} & \pi \\ \hline f(x) & 1 & 1.366 & 0.366 & -1 \\ & y_{0} & y_{1} & y_{2} & y_{3} \\ \hline \end{array}
By Trapezoidal
\int_{0}^{\pi}(\sin x+\cos x) d x=\frac{\pi / 3}{2}(1+(-1)+
\begin{aligned}2(1.366+0.366)) & \\ =\frac{\pi}{3}(1.732)=1.8137\end{aligned}
\begin{aligned}\int_{0}^{\pi}(\sin x+\cos x) d x&=\int_{0}^{\pi / 2}(\sin x+\cos x) d x \\&+\int_{\pi / 2}^{\pi}(\sin x+\cos x) d x\\ =&\left.(-\cos x+\sin x)\right|_{0} ^{\pi / 2}\\&+\left.(-\cos x+\sin x)\right|_{\pi / 2} ^{\pi}\\ =&[(0+1)-(-1+0)]\\&+[(1+0)-(0+1)] | x \\ =& 1+1+1-1 \\ =& 2 \end{aligned}
\begin{aligned} \text{Error}=& \text { Exact value - approx value } \\=& 2-1.8137 \\=& 0.1863 \end{aligned}
By Trapezoidal
\int_{0}^{\pi}(\sin x+\cos x) d x=\frac{\pi / 3}{2}(1+(-1)+
\begin{aligned}2(1.366+0.366)) & \\ =\frac{\pi}{3}(1.732)=1.8137\end{aligned}
\begin{aligned}\int_{0}^{\pi}(\sin x+\cos x) d x&=\int_{0}^{\pi / 2}(\sin x+\cos x) d x \\&+\int_{\pi / 2}^{\pi}(\sin x+\cos x) d x\\ =&\left.(-\cos x+\sin x)\right|_{0} ^{\pi / 2}\\&+\left.(-\cos x+\sin x)\right|_{\pi / 2} ^{\pi}\\ =&[(0+1)-(-1+0)]\\&+[(1+0)-(0+1)] | x \\ =& 1+1+1-1 \\ =& 2 \end{aligned}
\begin{aligned} \text{Error}=& \text { Exact value - approx value } \\=& 2-1.8137 \\=& 0.1863 \end{aligned}
Question 10 |
Numerical integration using trapezoidal rule gives the best result for a single variable function
linear | |
parabolic | |
logarithmic | |
hyperbolic |
Question 10 Explanation:
Trapezoidal rule gives the best result in single variable function when the function is linear (degree 1)
There are 10 questions to complete.
Reconstruct the question number 27….