# Find the area under the curve y=2x-3 from x=0 to x=6 using the right endpoints of 4 subintervals...

## Question:

Find the area under the curve {eq}y=2x-3 {/eq} from {eq}x=0 {/eq} to {eq}x=6 {/eq} using the right endpoints of 4 subintervals of equal width.

## Area Under the Curve:

Using the right endpoints of 4 sub-intervals of equal width, area is given by {eq}R_4=\sum_{i=1}^{4}f(x_i)\bigtriangleup x {/eq}

{eq}\bigtriangleup x=\frac{6-0}{2} {/eq}

Take {eq}x_0=0\,,,x_1=\frac{3}{2}\,,\,x_2=3\,,\,x_3=\frac{9}{2}\,,\,x_4=6 {/eq}

Put these values in {eq}f(x_i)=2x_i-3 {/eq} to find {eq}f(x_0)\,,\,f(x_1)\,,,f(x_2)\,,\,f(x_3)\,,\,f(x_4) {/eq}

Put values of {eq}\bigtriangleup x\,,\,f(x_0)\,,\,f(x_1)\,,,f(x_2)\,,\,f(x_3)\,,\,f(x_4) {/eq} in {eq}R_4=\sum_{i=1}^{4}f(x_i)\bigtriangleup x {/eq} to get the final answer.

{eq}\bigtriangleup x=\frac{6-0}{2}=\frac{3}{2} {/eq}

{eq}x_0=0\,,x_1=0+\frac{3}{2}=\frac{3}{2}\,,\,x_2=\frac{3}{2}+\frac{3}{2}=3\,,\,x_3=3+\frac{3}{2}=\frac{9}{2}\,,\,x_4=\frac{9}{2}+\frac{3}{2}=6 {/eq}

{eq}f(0)=-3\\ f\left ( \frac{3}{2} \right )=2\left ( \frac{3}{2} \right )-3=3-3=0\\ f(3)=2(3)-3=3\\ f\left ( \frac{9}{2} \right )=2\left ( \frac{9}{2} \right )-3=9-3=6\\ f(6)=2(6)-3=9 {/eq}

To find: {eq}R_4=\sum_{i=1}^{4}f(x_i)\bigtriangleup x {/eq}

{eq}R_4=\sum_{i=1}^{4}f(x_i)\bigtriangleup x\\ =0\left ( \frac{3}{2} \right )+3\left ( \frac{3}{2} \right )+6\left ( \frac{3}{2} \right )+9\left ( \frac{3}{2} \right )\\ =\frac{9}{2}+9+\frac{27}{2}\\ =18+9\\ =27 {/eq}