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  • Find the area of the region bounded by the curve y = x3 and y = x + 6 and x = 0.

Find the area of the region bounded by the curve y = x^3 and y = x + 6 and x = 0.

Answers (1)

Let’s start with a rough plot of the curve y = x\textsuperscript{3} along with the lines y = x + 6 and x = 0 

When x = 0, it means Y axis 

Questions says to find the area between curve and the line and Y axis 

First solve the y = x + 6 and  y = x\textsuperscript{3}, in order to find the interaction point 
\\\text { Put } y=x^{3} \text { in } y=x+6 \\ \Rightarrow x^{3}=x+6 \\ \Rightarrow x^{3}-x-6=0 \\ \\

 

For checking is 0, 1, 2 satisfy this cubic, shows 2 is one factor, therefore x – 2 is a factor. 

 

Solving the equation, 


\\ \Rightarrow(x-2)\left(x^{2}+2 x+3\right)=0 \\ \text{Observe that } x^{2}+2 x+3 \text{ don't have real roots} \\

Therefore, x = 2 

Substituting this x = 2 in y = x + 6, y = 8

Therefore, curves intersect at (2, 8) 

 

The area bounded will be 

 

item Area bounded = area by y = x\textsuperscript{3} on Y axis – area by y = x + 6 on Y axis   \ldots (1) \\

 

 

 


For finding area under y = x\textsuperscript{3} \\

 


\Rightarrow x=\sqrt[3]{y} \\

Integrate the equation from 0 to 8 
\\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\int_{0}^{8} \mathrm{y} \frac{1}{3} \mathrm{dy} \\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\left[\frac{\mathrm{y}^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]_{0}^{8} \\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\left[\frac{\mathrm{y}^{\frac{4}{3}}}{\frac{4}{3}}\right]_{0}^{8} \\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\frac{3}{4}\left[\mathrm{y}^{\frac{4}{3}}\right]_{0}^{8} \\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\frac{3}{4}[8 \overline{4/3}-0] \\ \\ \\\\\Rightarrow \int_{0}^{8} x d y=\frac{3}{4}\left(2^{3}\right)^{\frac{4}{3}} \\ \Rightarrow \int_{0}^{8} x d y=\frac{3}{4} 2^{4} \\ \Rightarrow \int_{0}^{8} x d y=\frac{3}{4} 16 \\ \Rightarrow \int_{0}^{8} x d y=12 \\ \\

 

Now, finding area under y = x + 6 

 

For finding the area from 6 to 8 because line passes through Y axis at 6 and extends upto 8, the point where curve and line intersects. 

 

           \item X = y - 6 \\

Integrating from 6 to 8 
\\ \Rightarrow \int_{6}^{8} x d y=\int_{6}^{8}(y-6) d y \\ \Rightarrow \int_{6}^{8} x d y=\left[\frac{y^{2}}{2}-6 y\right]_{6}^{8} \\ \Rightarrow \int_{6}^{8} x d y=\left[\left(\frac{8^{2}}{2}-6(8)\right)-\left(\frac{6^{2}}{2}-6(6)\right)\right] \\ \Rightarrow \int_{6}^{8} x d y=[32-48-18+36] \\ \Rightarrow \int_{6}^{8} x d y=2 \\ \\

Using the equation (1) 

Area bound was found to be 12 – 2 = 10 unit\textsuperscript{2 }\\

 

Therefore, the area was found to be 10 unit\textsuperscript{2 }\\

 

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