Tag: the trapezium rule

Questions Related to the trapezium rule

The second approximation of roots of $x^3-5x-7=0$ in the interval $(2,3)$ by the method of false position is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $1.735$

  2. $2.375$

  3. $3.735$

  4. $2.735$

Show answer
Correct Option: D
Explanation:

Here, $x^3-5x-7=0$

Let $f(x)=x^3-5x-7$
First Iteration:
Here, $f(2)=-9<0$ and $f(3)=5>0$
Now, Root lies between $x _0=2$ and $x _1=3$
$x _2=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2-(-9)\times \dfrac{3-2}{5-(-9)}=2.64286$
Second Iteration:
Here, $f(2.64286)=-1.75474$ and $f(3)=5>0$

Now, Root lies between $x _0=2.64286$ and $x _1=3$
$x _3=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2.64-(-1.75)\times \dfrac{3-2.64}{5-(-1.75)}=2.73564$

The third approximation of root of $x^3-x^2-1=0$ in the interval $(1,2)$ using successive bisection method is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $1.475$

  2. $1.375$

  3. $2.213$

  4. $1.564$

Show answer
Correct Option: B
Explanation:

We have to find the third approximation of root of the equation $x^3-x^2-1=0$ in the interval $(1,2)$ using successive Bisection method.

$\textbf{Iteration 1: k=0}$

$c _0=\dfrac{a _0+b _0}{2}=\dfrac{1+2}{2}=1.5$

Since $f(c _0)f(a _0)=f(1.5)f(1)<0$

Therefore set $a _1=a _0,b _1=c _0$

$\textbf{Iteration 2: k=1}$

$c _1=\dfrac{a _1+b _1}{2}=\dfrac{1+1.5}{2}=1.25$

Since $f(c _1)f(a _1)=f(1.25)f(1)>0$

Therefore set $a _2=c _1,b _2=b _1$

$\textbf{Iteration 3: k=2}$

$c _2=\dfrac{a _2+b _2}{2}=\dfrac{1.25+1.5}{2}=1.375$

Thus the third approximation of the root is $1.375$ respectively.

By successive bisection method, the cube root of $2$ between the interval (1,1.5)_is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $1.2813$

  2. $1.2121$

  3. $1.013$

  4. $1.475$

Show answer
Correct Option: A
Explanation:

Function can be written as $f(x)=x^3-2$


First Iteration:
$f(1)=-1<0$ and $f(1.5)=1.375>0$
Now, root lies between $1$ and $1.5$
So,
$x _0=\dfrac{1+1.5}{2}=1.25$
$f(x _0)=-0.04688<0$

Second Iteration:

$f(1.25)=-0.04688<0$ and $f(1.5)=1.375>0$
Now, root lies between $1.25$ and $1.5$
So,
$x _1=\dfrac{1.25+1.5}{2}=1.375$

$f(x _1)=0.59961>0$

Third Iteration:

$f(1.25)=-0.04688<0$ and $f(1.375)=0.59961>0$
Now, root lies between $1.25$ and $1.375$
So,
$x _2=\dfrac{1.25+1.375}{2}=1.3125$
$f(x _2)=0.26099>0$

Fourth Iteration:
$f(1.25)=-0.04688<0$ and $f(1.3125)=0.26099>0$
Now, root lies between $1.25$ and $1.3125$
So,
$x _2=\dfrac{1.25+1.3125}{2}=1.28125\approx 1.2813$

The value of $\cdot4267\ E\ 10 \div \cdot2437\ E -02.$ is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $\cdot1751\ E\ 03$

  2. $\cdot1752\ E\ 13$

  3. $\cdot1751\ E\ 13$

  4. $\cdot1762\ E\ 13$

Show answer
Correct Option: C
Explanation:

$0.4267:E:10\div0.2437 : E :-02=?$

The Scientific format displays a number in exponential notation, replacing part of the number with $E+n$, where $E$ (stands for exponent) multiplies the preceding number by $10$ to the $n^{th}$ power.

That is $1.23E+10$ can be written as $1.23 \times 10^{10}$

$0.4267:E:10\div0.2437 : E :-02=\dfrac{0.4267 \times 10^{10}}{0.2437 \times 10^{-2}}$

                                                       $= 1.75092 \times 10^{12}$

                                                       $= 1.751 \times 10^{12}$

                                                       $= .1751 \times 10^{13}$

                                                       $= 0.1751 : E :13$

Hence $0.4267:E:10\div0.2437 : E :-02=0.1751 : E :13$

The third approximation of roots of $x^3-9x+1=0$ in the interval $(2,4)$ by the method of false position is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $8.23$

  2. $1.25$

  3. $2.85$

  4. $2.12$

Show answer
Correct Option: C
Explanation:

Here, $x^3-9x+1=0$

Let $f(x)=x^3-9x+1$
First Iteration:
Here, $f(2)=-9<0$ and $f(4)=29>0$
Now, Root lies between $x _0=2$ and $x _1=4$
$x _2=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2-(-9)\times \dfrac{4-2}{29-(-9)}=2.47368$
Second Iteration:
Here, $f(2.47368)=-6.1264$ and $f(2)=29>0$

Now, Root lies between $x _0=2.47368$ and $x _1=4$
$x _3=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2.47-(-6.13)\times \dfrac{4-2.47}{29-(-6.13)}=2.73989$

Third Iteration:

Here, $f(2.73989)=-3.09067$ and $f(4)=29>0$
Now, Root lies between $x _0=2.73989$ and $x _1=4$
$x _4=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2.74-(-3.09)\times \dfrac{4-2.74}{29-(-3.09)}=2.86125$

If the error committed in measuring the radius of the circle is $0.05\%$, then the corresponding error in calculating the area is:

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $0.05\%$

  2. $0.025\%$

  3. $0.25\%$

  4. $0.1\%$

Show answer
Correct Option: D
Explanation:

$\dfrac { dr }{ r } =0.05\Rightarrow dr=(0.05)r$

Area of circle $=\pi r^2$
$\ A=\pi r^{ 2 }\Rightarrow \dfrac { dA }{ dr } =2\pi r\ dA=2\pi rdr\Rightarrow \dfrac { dA }{ A } =\dfrac { 2\pi rdr }{ \pi r^{ 2 } } =\dfrac { 2dr }{ r } \ \therefore \dfrac { dA }{ A } =2(0.05)^{ 2 }=0.1$
$\therefore$ Corresponding error in area $= 0.1\%$