Question

hola, me podrian ayudar? es mi tarea y es para hoy en la noche :(((
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Answer 1

$\int ^{4} _{ - 2} \: {x}^{3} dx = \frac{ {x}^{4} }{4} \mid^{ \: \: \: 4} _{ - 2} \\ = \frac{ {4}^{4} }{4} - \frac{ {( - 2)}^{4} }{4} \\ = 64 - 4 \\ = 60$

$\int ^{1} _{ - 1}2 {x}^{4} dx = \frac{2 {x}^{5} }{5} \mid ^{ \: \: \: 1} _{ - 1} \\ = \frac{2 \times ({1})^{5} }{5} - \frac{2 \times ( - 1) ^{5} }{5} \\ = \frac{2}{5} - ( - \frac{2}{5} ) \\ = \frac{2}{5} + \frac{2}{5} \\ = \frac{4}{5}$

$\int ^{2} _{ -3} - 3x + 4dx = \\ - \frac{3 {x}^{2} }{2} + 4x\mid ^{ \: \: \: 2} _{ - 3} \\ = ( - \frac{3 \times {2}^{2} }{2} + 4 \times 2) - ( - \frac{3 \times ( - 3)^{2} }{2 } + 4 \times ( - 3)) \\ = ( - 6 + 8) - ( - \frac{27}{2} - 12) \\ = 2 + 12 + \frac{27}{2} \\ = 14 + \frac{27}{2} \\ = \frac{55}{2}$

$\int ^{2} _{ 0} \sqrt{x} \: dx = \\ \frac{2 \sqrt{ {x}^{3} } }{3} \mid^{2} _{ 0} \\ = \frac{2 \sqrt{ {2}^{3} } }{3} - \frac{2 \sqrt{ {0}^{3} } }{3} \\ = \frac{4 \sqrt{2} }{3}$

$\int ^{2} _{ - 2}4x + 1 \: dx = \\ 4 {x}^{2} + x\mid ^{ \: \: \: 2} _{ - 2} \\ = (4 \times (2) ^{2} + 2) - (4 \times ( - 2) ^{2} + ( - 2)) \\ = (4 \times 4 + 2) - (4 \times 4 - 2) \\ = (16 + 2) - (16 - 2) \\ = 18 - 14 \\ = 4$