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How do you evaluate the definite integral #int(x^3-x^2+1)dx# from #[-1,2]#?

Answer 1

#15/4#

Use the #color(blue)"power rule for integration"#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(intax^ndx=a/(n+1)x^(n+1))color(white)(2/2)|)))#

Apply this to each term.

#rArrint_-1^2(x^3-x^2+1)dx#

#=[1/4x^4-1/3x^3+x]_-1^2#

#=[1/4(2)^4-1/3(2)^3+2]-[1/4(-1)^4-1/3(-1)^3-1]#

#=(4-8/3+2)-(1/4+1/3-1)#

#=10/3-(-5/12)=15/4#

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Answer 2

Paisley Johnson

To evaluate the definite integral of ( \int_{-1}^{2} (x^3 - x^2 + 1) , dx ) from -1 to 2, you would first find the antiderivative of the given function, which is ( \frac{1}{4}x^4 - \frac{1}{3}x^3 + x ). Then, you would substitute the upper limit (2) into this antiderivative and subtract the result from the substitution of the lower limit (-1). This yields the value of the definite integral.

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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.