Pearl Carney

2021-11-20

What is the integral of the functions:
a) ${\int }_{-3}^{-2}\left(2{y}^{2}+12y+19\right)dy$
b) ${\int }_{-3}^{-2}\left(\frac{-{y}^{2}}{2}-4y-10\right)dy$

Sact1989

Given integrals are
a) ${\int }_{-3}^{-2}\left(2{y}^{2}+12y+19\right)dy$
b) ${\int }_{-3}^{-2}\left(\frac{-{y}^{2}}{2}-4y-10\right)dy$
On solving the first integral
${\int }_{-3}^{-2}\left(2{y}^{2}+12y+19\right)dy$
$={\left[2\frac{{y}^{2+1}}{2+1}+12\frac{{y}^{1+1}}{1+1}+19y\right]}_{-3}^{-2}$
$={\left[2\frac{{y}^{3}}{3}+12\frac{{y}^{2}}{2}+19y\right]}_{-3}^{-2}$
$=\left[2\frac{{\left(-2\right)}^{3}}{3}+12\frac{{\left(-2\right)}^{2}}{2}+19\left(-2\right)-\left(2\frac{{\left(-3\right)}^{3}}{3}+12\frac{{\left(-3\right)}^{2}}{2}+19\left(-3\right)\right)\right]$
$=\left[\frac{16}{3}+24-38-\left(-18+54-57\right)\right]$
$=\left[\frac{-58}{3}+21\right]$
$=\frac{5}{3}$

Ruth Phillips

On solving the second integral
${\int }_{-3}^{-2}\left(\frac{-{y}^{2}}{2}-4y-10\right)dy$
$={\left[-\frac{{y}^{2+1}}{2\left(2+1\right)}-4\frac{{y}^{1+1}}{1+1}-10y\right]}_{-3}^{-2}$
$={\left[-\frac{{y}^{3}}{6}-4\frac{{y}^{2}}{2}-10y\right]}_{-3}^{-2}$
$=\left[-\frac{{\left(2\right)}^{3}}{6}-4\frac{{\left(-2\right)}^{2}}{2}-10\left(-2\right)-\left(-\frac{{\left(-3\right)}^{3}}{6}-4\frac{{\left(-3\right)}^{2}}{2}-10\left(-3\right)\right)\right]$
$=\left[\frac{4}{3}-8+20-\left(\frac{9}{2}-18+30\right)\right]$
$=\left[\frac{40}{3}-\frac{33}{2}\right]$
$=\frac{-19}{6}$

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