Find the area enclosed between the curve \(\displaystyle y = - 4x - x^2\) and the line \(\displaystyle y = - \frac{1}{4}x\)
To find the point of intersection:
\(\displaystyle - 4x - x^2 = - \frac{1}{4}x\)
\(\displaystyle x = - \frac{{15}}{4}\)
Limits are \(\displaystyle x = - \frac{{15}}{4}{\rm and }x = 0\)
\(\displaystyle \int_{ - {\textstyle{{15} \over 4}}}^0 {( - 4x - x^2 )dx = - 2} \left[ {x^2 } \right]_{ - {\textstyle{{15} \over 4}}}^0 - \frac{1}{3}\left[ {x^3 } \right]_{ - {\textstyle{{15} \over 4}}}^0\)
\(\displaystyle = - 2\left[ { - 14.06} \right] - \frac{1}{3}\left[ {52.70} \right] = 10.547\)
\(\displaystyle \int_{ - {\textstyle{{15} \over 4}}}^0 {( - \frac{1}{4}x)dx = - \frac{1}{8}} \left[ {x^2 } \right]_{ - {\textstyle{{15} \over 4}}}^0 = \left[ { - 14.06} \right] = 1.757\)
So area enclosed is
\(\displaystyle 10.247 - 1.757 = 8.79\)
Is this right?
Thanks
To find the point of intersection:
\(\displaystyle - 4x - x^2 = - \frac{1}{4}x\)
\(\displaystyle x = - \frac{{15}}{4}\)
Limits are \(\displaystyle x = - \frac{{15}}{4}{\rm and }x = 0\)
\(\displaystyle \int_{ - {\textstyle{{15} \over 4}}}^0 {( - 4x - x^2 )dx = - 2} \left[ {x^2 } \right]_{ - {\textstyle{{15} \over 4}}}^0 - \frac{1}{3}\left[ {x^3 } \right]_{ - {\textstyle{{15} \over 4}}}^0\)
\(\displaystyle = - 2\left[ { - 14.06} \right] - \frac{1}{3}\left[ {52.70} \right] = 10.547\)
\(\displaystyle \int_{ - {\textstyle{{15} \over 4}}}^0 {( - \frac{1}{4}x)dx = - \frac{1}{8}} \left[ {x^2 } \right]_{ - {\textstyle{{15} \over 4}}}^0 = \left[ { - 14.06} \right] = 1.757\)
So area enclosed is
\(\displaystyle 10.247 - 1.757 = 8.79\)
Is this right?
Thanks