3 The diagram shows a sketch of part of the curve with equation y=9-3x-5x^2-x^3 and the line with equation y=4-4x The line cuts the curve at the points A(-1,8) and B(1,0) Find the area of the shaded region between AB and the curve.

## Step 1:<br />Firstly, we can calculate the area by integrating the height from the x-coordinate of point A to the x-coordinate of point B. We calculate the height by subtracting the y-value of the line at x from the y-value of the curve at x.<br />Formally, we have a height $H(x)=f(x)-g(x)$, where $f(x)$ is the y-value of the curve at x and $g(x)$ = y-value of the line at x. From the given problem we know \(f(x) = 9 - 3x - 5x^{2} -x^{3}\) and \(g(x) = 4 - 4x\) <br /><br />## Step 2:<br />Now by simplicity and commutative rules and due to \(H(x)\) being a differene, we can rearrange to get \(H(x) = (-x^{3} - 5x^{2} + x + 5)\)<br /><br />## Step 3: <br />Finally, we use an integral to calculate the area as follows for x values within the limits of A and B which are -1 and 1 respectively.<br /><br />### \[A = \int_{{-1}}^{1}H(x)dx\]<br /><br />## Step 4: <br />Now, integral of \(H(X)\) within the limits of -1 and 1 gets us to;<br /><br />### \[A = \left[ - \frac { x^{4} } { 4 } - \frac { 5x^{3} } { 3 } + \frac { x^{2} } { 2 } + 5x \right] _{-1} ^{1}\]<br /><br />## Step 5: <br />Proceeding with above gets as the result for the integral to be<br /><br />### \[A = \left( - \frac { 1 } { 4 } - \frac { 5 } { 3 } + \frac { 1 } { 2 } + 5 \right) - \left( - \frac { 1 } { 4 } + \frac { 5 } { 3 } + \frac { 1 } { 2 } - 5 \right)\]<br /><br />## Step 6:<br />Further simplification turns into:<br /><br />### \[A= \frac { 2 } { 3 } - 5 = \frac { 20 } { 15 } - \frac { 75 } { 15 } = -\frac { 55 } { 15 } = \frac { 11 } { -3 } = - \frac { 11 } { 3 }\]