What is the integral of the functions: a) ∫−3−2(2y2+12y+19)dy b) ∫−3−2(−y22−4y−10)dy

Given integrals are a) ∫−3−2(2y2+12y+19)dy b) ∫−3−2(−y22−4y−10)dy On solving the first integral ∫−3−2(2y2+12y+19)dy =[2y2+12+1+12y1+11+1+19y]−3−2 =[2y33+12y22+19y]−3−2 =[2(−2)33+12(−2)22+19(−2)−(2(−3)33+12(−3)22+19(−3))] =[163+24−38−(−18+54−57)] =[−583+21] =53

On solving the second integral ∫−3−2(−y22−4y−10)dy =[−y2+12(2+1)−4y1+11+1−10y]−3−2 =[−y36−4y22−10y]−3−2 =[−(2)36−4(−2)22−10(−2)−(−(−3)36−4(−3)22−10(−3))] =[43−8+20−(92−18+30)] =[403−332] =−196

Given integrals are a) ∫−3−2(2y2+12y+19)dy b) ∫−3−2(−y22−4y−10)dy On solving the first integral ∫−3−2(2y2+12y+19)dy =[2y2+12+1+12y1+11+1+19y]−3−2 =[2y33+12y22+19y]−3−2 =[2(−2)33+12(−2)22+19(−2)−(2(−3)33+12(−3)22+19(−3))] =[163+24−38−(−18+54−57)] =[−583+21] =53

On solving the second integral ∫−3−2(−y22−4y−10)dy =[−y2+12(2+1)−4y1+11+1−10y]−3−2 =[−y36−4y22−10y]−3−2 =[−(2)36−4(−2)22−10(−2)−(−(−3)36−4(−3)22−10(−3))] =[43−8+20−(92−18+30)] =[403−332] =−196

Answered: "What is the integral of the functions: a)..."

Pearl Carney

Answered question

2021-11-20

What is the integral of the functions:
a)

\int_{- 3}^{- 2} (2 y^{2} + 12 y + 19) d y

\int_{- 3}^{- 2} (\frac{- y^{2}}{2} - 4 y - 10) d y

Answer & Explanation

Sact1989

Beginner2021-11-21Added 10 answers

Given integrals are
a)

\int_{- 3}^{- 2} (2 y^{2} + 12 y + 19) d y

\int_{- 3}^{- 2} (\frac{- y^{2}}{2} - 4 y - 10) d y

On solving the first integral

\int_{- 3}^{- 2} (2 y^{2} + 12 y + 19) d y

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

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

Ruth Phillips

Beginner2021-11-22Added 18 answers

On solving the second integral

\int_{- 3}^{- 2} (\frac{- y^{2}}{2} - 4 y - 10) d y

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

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

= [- \frac{{(2)}^{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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