What is $\int_{1}^{4} (.2 x^{3} - 2 x + 4) d x$ $int_1^4(.2x^3-2x+4)dx$ ?

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Find $\int_{1}^{4} (.2 x^{3} - 2 x + 4) d x$ $int_1^4(.2x^3-2x+4)dx$

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Answer 1 · 2018-03-16T18:06:22

$124.5$ $124.5$

$\int_{1}^{4} (2 x^{3} - 2 x + 4) d x$ $int_1^4 (2x^3-2x+4) dx$

= $[(\frac{2 x^{4}}{4}) - (\frac{2 x^{2}}{2}) + 4 x]$ $[((2x^4)/4)-((2x^2)/2)+4x]$ With upper limit x=4 and lower limit x=1

Apply your limits in the integrated expression, i.e subtract your lower limit from your upper limit.

$= (128 - 16 - 16) - ((\frac{1}{2}) - 1 + 4)$ $=(128-16-16)-((1/2)-1+4)$

$= 128 - 3 (\frac{1}{2}) = 124.5$ $=128-3(1/2)=124.5$