Question

Find the area under the curve $y=\sqrt{6x+4}$ (above x-axis) from x= 0 to x=2.

Answer 1

Hint: Plot the graph of the curve. Identify the region whose area is to be found. Use the fact that the area bounded by the curve y=f(x), the x-axis and the ordinates x=a and x=b is given by $A=\int_{a}^{b}{\left| f\left( x \right) \right|dx}$. Hence prove that the required area is given by $A=\int_{0}^{2}{\sqrt{6x+4}dx}$. Evaluate the integral and hence find the area.

Complete step-by-step answer:

As is evident from the graph, we need to find the area of the region BCEDB.
We know that the area bounded by the curve y=f(x), the x-axis and the ordinates x=a and x=b is given by $A=\int_{a}^{b}{\left| f\left( x \right) \right|dx}$.
Hence the area bounded by $y=\sqrt{6x+4}$, the x-axis and the ordinates x= 0 and x= 2 is given by
$A=\int_{0}^{2}{\left| \sqrt{6x+4} \right|dx}$
Since $\left| \sqrt{x} \right|=\sqrt{x},$ we get
$A=\int_{0}^{2}{\sqrt{6x+4}dx}$
Put $\sqrt{6x+4}=t\Rightarrow 6x+4={{t}^{2}}$
Differentiating with respect to t, we get
$6dx=dt\Rightarrow dx=\dfrac{1}{6}2tdt=\dfrac{tdt}{3}$
When x= 0, we have $t=\sqrt{6\left( 0 \right)+4}=\sqrt{4}=2$
When x=2, we have $t=\sqrt{6\left( 2 \right)+4}=\sqrt{12+4}=4$
Hence, we have
$A=\int_{2}^{4}{{{t}^{2}}\dfrac{dt}{3}}$
We know that $\int_{a}^{b}{kf\left( x \right)dx}=k\int_{a}^{b}{f\left( x \right)dx}$
Hence, we have
$A=\dfrac{1}{3}\int_{2}^{4}{{{t}^{2}}dt}=\dfrac{1}{3}\left( \left. \dfrac{{{t}^{3}}}{3} \right|_{2}^{4} \right)=\dfrac{1}{9}\left( 64-8 \right)=\dfrac{56}{9}$
Hence the area bounded by the curve $y=\sqrt{6x+4}$, the x-axis and the ordinates x= 0 and x =2 is 2 square units.

Note: We can solve the integral $A=\int_{0}^{2}{\sqrt{6x+4}dx}$ directly by using the property $\int{{{\left( ax+b \right)}^{n}}dx=\dfrac{{{\left( ax+b \right)}^{n+1}}}{n+1}}+C$
Hence, we have
$\int_{0}^{2}{\sqrt{6x+4}dx}=\left. \dfrac{2}{3}\times {{\left( 6x+4 \right)}^{\dfrac{3}{2}}}\times \dfrac{1}{6} \right|_{0}^{2}=\dfrac{1}{9}\left( 64-8 \right)=\dfrac{56}{9}$, which is the same as obtained above.