Pearl Carney

2021-11-20

What is the integral of the functions:

a)${\int}_{-3}^{-2}(2{y}^{2}+12y+19)dy$

b)${\int}_{-3}^{-2}(\frac{-{y}^{2}}{2}-4y-10)dy$

a)

b)

Sact1989

Beginner2021-11-21Added 10 answers

Given integrals are

a)${\int}_{-3}^{-2}(2{y}^{2}+12y+19)dy$

b)${\int}_{-3}^{-2}(\frac{-{y}^{2}}{2}-4y-10)dy$

On solving the first integral

${\int}_{-3}^{-2}(2{y}^{2}+12y+19)dy$

$={[2\frac{{y}^{2+1}}{2+1}+12\frac{{y}^{1+1}}{1+1}+19y]}_{-3}^{-2}$

$={[2\frac{{y}^{3}}{3}+12\frac{{y}^{2}}{2}+19y]}_{-3}^{-2}$

$=[2\frac{{(-2)}^{3}}{3}+12\frac{{(-2)}^{2}}{2}+19(-2)-(2\frac{{(-3)}^{3}}{3}+12\frac{{(-3)}^{2}}{2}+19(-3))]$

$=[\frac{16}{3}+24-38-(-18+54-57)]$

$=[\frac{-58}{3}+21]$

$=\frac{5}{3}$

a)

b)

On solving the first integral

Ruth Phillips

Beginner2021-11-22Added 18 answers

On solving the second integral

${\int}_{-3}^{-2}(\frac{-{y}^{2}}{2}-4y-10)dy$

$={[-\frac{{y}^{2+1}}{2(2+1)}-4\frac{{y}^{1+1}}{1+1}-10y]}_{-3}^{-2}$

$={[-\frac{{y}^{3}}{6}-4\frac{{y}^{2}}{2}-10y]}_{-3}^{-2}$

$=[-\frac{{\left(2\right)}^{3}}{6}-4\frac{{(-2)}^{2}}{2}-10(-2)-(-\frac{{(-3)}^{3}}{6}-4\frac{{(-3)}^{2}}{2}-10(-3))]$

$=[\frac{4}{3}-8+20-(\frac{9}{2}-18+30)]$

$=[\frac{40}{3}-\frac{33}{2}]$

$=\frac{-19}{6}$

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