Question

Sketch the region lying in the first quadrant and bounded by \[y = 9{x^2},\;x = 0,\;y = 1\] and \[y = 4\] Find the area of the region, using integration.

Answer 1

Hint: Before we begin, we must understand that the y-axis is also known as the x=0 line, hence the y-axis is being discussed in the question. Use the knowledge that gives the area limited by the curve \[y = f\left( x \right)\] the \[x\] -axis, and the ordinates \[x = a\] and \[x = b\].

Formula Used: $\int {{x}^{n+1}}dx=\dfrac{{{x}^{n+1}}}{n+1}$
Area of triangle = $\dfrac{1}{2}\times base\times height$

Complete step by step Solution:
As a result, in this question, we are asked to discover the area limited by the curve \[y = 9{x^2}\] and the \[y\] -axis, which means the region between the abscissas \[y = a\] and \[y = b\] and we can exploit this information as follows.
The equation that determines the region enclosed by the curve\[x = f\left( y \right)\], the \[y\]-axis, and the abscissas \[y = a\] and \[y = b\] is

Now, let us write the curve function as
\[ \Rightarrow y = \left( {9{x^2}} \right)\]
Now, we have to calculate for \[{x^2}\] we get,
\[ \Rightarrow {x^2} = \dfrac{y}{9}\]
Now, let’s square the above equation on both sides, we get
\[ \Rightarrow x = \pm \dfrac{{\sqrt y }}{3}\]
Now, it can be also written as,
\[ \Rightarrow f(y) = \pm \dfrac{{\sqrt y }}{3}\]
Now that we have a better understanding of the figure and the fact that the first quadrant must be taken into account, we can simply assume that the value of \[f\left( y \right)\] is positive.
Hence, the required area is obtained as,
\[\smallint _a^bf(y)dx\]
Now, the area of the region encompassed by the curve must be determined, as stated in the question.
As a result,\[y\] has values of \[1\] and \[4\] which correspond to \[a\] and \[b\] respectively.
To obtain the requisite area of the region, we can now easily form the integral as shown below.
Area\[ = \int_a^b f (y)dy\]
\[ = \int_1^4 {\dfrac{{\sqrt {(y)} }}{3}} dy\]
Now we have to take the constant out by applying linearity, we get
\[ = \dfrac{1}{3} \cdot \int_1^4 {\sqrt y } dy\]
Let us apply the power rule, we obtain
\[ = \dfrac{1}{3}\left[ {\dfrac{2}{3}{y^{\dfrac{3}{2}}}} \right]_1^4\]
Now, we have to compute the boundaries:
\[ = \dfrac{1}{3}\left[ {\dfrac{{{{(4)}^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}} - \dfrac{{{{(1)}^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}}} \right]\]
In simplification, we get
\[ = \left[ {\dfrac{{2\left( {{{(2)}^{\dfrac{3}{2} \times 2}} - 1} \right)}}{9}} \right]\]
\[ = \left[ {\dfrac{{2\left( {{{(2)}^3} - 1} \right)}}{9}} \right]\]
On further simplification, we obtain
\[ = \left[ {\dfrac{{14}}{9}} \right]\]
Hence, we have total area\[ = \dfrac{{14}}{9}{\rm{units }}\].

Note:If the students don't understand the fundamentals of integration, they risk making a mistake because, without them, it would be impossible to arrive at the right solution. It is crucial to know the answer to the following question in advance. The integral of \[{x^n}\] is defined as\[\smallint {x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\].