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How do you evaluate the definite integral #int (-3v+4)dv# from [2,5]?

Answer 1

Luke Alvarez

#-39/2#

#int_2^5(-3v+4)dv#

the antiderivative is #F(v)=-3/2v^2+4v#

simply evaluate #F(5)-F(2)# #=-3/2(5)^2+4(5)+3/2(2)^2-4(2)# #=-75/2+20+6-8# #=-39/2#

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Answer 2

Emily Ammerman

To evaluate the definite integral of (\int_{2}^{5} (-3v+4) , dv), you first need to find the antiderivative of the integrand, which is (-\frac{3}{2}v^2 + 4v). Then, you evaluate this antiderivative at the upper and lower bounds of integration and subtract the lower value from the upper value.

[\left[-\frac{3}{2}v^2 + 4v\right]_{2}^{5}]

[= \left(-\frac{3}{2}(5)^2 + 4(5)\right) - \left(-\frac{3}{2}(2)^2 + 4(2)\right)]

[= \left(-\frac{3}{2}(25) + 20\right) - \left(-\frac{3}{2}(4) + 8\right)]

[= \left(-\frac{75}{2} + 20\right) - \left(-6 + 8\right)]

[= (-\frac{75}{2} + 20) - (-6 + 8)]

[= (-\frac{75}{2} + 20) - 2]

[= -\frac{75}{2} + 20 - 2]

[= -\frac{75}{2} + 18]

[= -\frac{75}{2} + \frac{36}{2}]

[= -\frac{39}{2}]

So, (\int_{2}^{5} (-3v+4) , dv = -\frac{39}{2}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.