Solveeit Logo
Login

Question

Question: How do you use the trapezoidal rule with \(n=2\) to approximate the area under the curve \(y=\dfrac{...

How do you use the trapezoidal rule with n=2n=2 to approximate the area under the curve y=1x2y=\dfrac{1}{{{x}^{2}}} from 11 to 33?

Explanation

Solution

In the question, we have to find the area under the curve y=1x2y=\dfrac{1}{{{x}^{2}}} from 11 to 33 using the trapezoidal rule. Firstly, we have to calculate the step-size by using the formula h=b−anh=\dfrac{b-a}{n}, where n is given in the question as n=2n=2, and a and b are the lower and the upper limits of the integration. With the help of the step size value, we can calculate x0,x1,x2{{x}_{0}},{{x}_{1}},{{x}_{2}}, where x0=a{{x}_{0}}=a and xi=a+ih{{x}_{i}}=a+ih. From these values of xn{{x}_{n}}, we can calculate yn{{y}_{n}} using yn=f(xn){{y}_{n}}=f\left( {{x}_{n}} \right). Then finally the value of the integral will be calculated by using the formula ∫abf(x)dx=h(y0+yn2+y1+y2+.....+yn)\int_{a}^{b}{f\left( x \right)dx}=h\left( \dfrac{{{y}_{0}}+{{y}_{n}}}{2}+{{y}_{1}}+{{y}_{2}}+.....+{{y}_{n}} \right).

Complete step-by-step answer:
We know that the area under a curve is calculated by integrating it with respect to x. According to the question, the function is
⇒y=1x2.......(i)\Rightarrow y=\dfrac{1}{{{x}^{2}}}.......\left( i \right)
The given limits are from 11 to 33. Therefore, the area under the given curve will be given by the integral
⇒A=∫131x2dx\Rightarrow A=\int_{1}^{3}{\dfrac{1}{{{x}^{2}}}dx}
Now, the step size is given by
⇒h=b−an\Rightarrow h=\dfrac{b-a}{n}
Substituting a=1a=1, b=3b=3 and n=2n=2, we get
⇒h=3−12 ⇒h=22 ⇒h=1......(iii) \begin{aligned} & \Rightarrow h=\dfrac{3-1}{2} \\\ & \Rightarrow h=\dfrac{2}{2} \\\ & \Rightarrow h=1......\left( iii \right) \\\ \end{aligned}
Now, we know that
⇒xi=a+ih\Rightarrow {{x}_{i}}=a+ih
Substituting i=0i=0 we get
⇒x0=a+(0)h ⇒x0=a ⇒x0=1 \begin{aligned} & \Rightarrow {{x}_{0}}=a+\left( 0 \right)h \\\ & \Rightarrow {{x}_{0}}=a \\\ & \Rightarrow {{x}_{0}}=1 \\\ \end{aligned}
Now, substituting i=1i=1 we get
⇒x1=a+(1)h ⇒x1=1+(1)(1) ⇒x1=2 \begin{aligned} & \Rightarrow {{x}_{1}}=a+\left( 1 \right)h \\\ & \Rightarrow {{x}_{1}}=1+\left( 1 \right)\left( 1 \right) \\\ & \Rightarrow {{x}_{1}}=2 \\\ \end{aligned}
Now substituting i=2i=2 we get
⇒x2=a+2h ⇒x2=1+2(1) ⇒x2=3 \begin{aligned} & \Rightarrow {{x}_{2}}=a+2h \\\ & \Rightarrow {{x}_{2}}=1+2\left( 1 \right) \\\ & \Rightarrow {{x}_{2}}=3 \\\ \end{aligned}
Now, the values of yi{{y}_{i}} are given by
⇒yi=f(xi) ⇒yi=1xi2 \begin{aligned} & \Rightarrow {{y}_{i}}=f\left( {{x}_{i}} \right) \\\ & \Rightarrow {{y}_{i}}=\dfrac{1}{{{x}_{i}}^{2}} \\\ \end{aligned}
Substituting i=0i=0 we get
⇒y0=1x02 ⇒y0=112 ⇒y0=1......(iv) \begin{aligned} & \Rightarrow {{y}_{0}}=\dfrac{1}{{{x}_{0}}^{2}} \\\ & \Rightarrow {{y}_{0}}=\dfrac{1}{{{1}^{2}}} \\\ & \Rightarrow {{y}_{0}}=1......\left( iv \right) \\\ \end{aligned}
Similarly, we get

& \Rightarrow {{y}_{1}}=\dfrac{1}{{{x}_{1}}^{2}} \\\ & \Rightarrow {{y}_{1}}=\dfrac{1}{{{2}^{2}}} \\\ & \Rightarrow {{y}_{1}}=\dfrac{1}{4} \\\ & \Rightarrow {{y}_{1}}=0.25......\left( v \right) \\\ \end{aligned}$$ And $\begin{aligned} & \Rightarrow {{y}_{2}}=\dfrac{1}{{{x}_{2}}^{2}} \\\ & \Rightarrow {{y}_{2}}=\dfrac{1}{{{3}^{2}}} \\\ & \Rightarrow {{y}_{2}}=\dfrac{1}{9} \\\ & \Rightarrow {{y}_{2}}=0.11......\left( vi \right) \\\ \end{aligned}$ Now, we know that the integration using the trapezoidal rule is given by $\Rightarrow \int_{a}^{b}{f\left( x \right)dx}=h\left( \dfrac{{{y}_{0}}+{{y}_{n}}}{2}+{{y}_{1}}+{{y}_{2}}+.....+{{y}_{n}} \right)$ In this case, $n=2$. On putting this above we get $\Rightarrow \int_{1}^{3}{\dfrac{1}{{{x}^{2}}}dx}=h\left( \dfrac{{{y}_{0}}+{{y}_{2}}}{2}+{{y}_{1}}+{{y}_{2}} \right)$ Substituting the equations (iii), (iv), (v), (vi) we get $\begin{aligned} & \Rightarrow \int_{1}^{3}{\dfrac{1}{{{x}^{2}}}dx}=1\left( \dfrac{1+0.11}{2}+0.25+0.11 \right) \\\ & \Rightarrow \int_{1}^{3}{\dfrac{1}{{{x}^{2}}}dx}=\left( \dfrac{1.11}{2}+0.25+0.11 \right) \\\ & \Rightarrow \int_{1}^{3}{\dfrac{1}{{{x}^{2}}}dx}=\left( 0.555+0.25+0.11 \right) \\\ & \Rightarrow \int_{1}^{3}{\dfrac{1}{{{x}^{2}}}dx}=0.915 \\\ \end{aligned}$ Hence, the area under the given curve is equal to $0.915$ square units. **Note:** The value of the integral calculated using the trapezoidal rule in the above solution may deflect much from the original value. On integrating the given function and substituting the limits, we will get the value equal to $0.67$. The reason is that the value of n, equal to two, is very much small. As the value of n is increased, the value of the integral using the trapezoidal rule will approach the original value.