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

Answer 1

Evelyn Agard

#2#

Given: #int(2x-3)dx# from #[1,3]#

#int_1^3(2x-3)dx =int_1^3(2x)dx - int_1^3 3 dx #

Use: #int(kx^n)dx = k*1/(n+1)x^(n+1) + C#

#int_1^3(2x)dx - int_1^3 3 dx = 2*1/2 x^2 - 3x |_1^3#

# = x^2 - 3x |_1^3 = (3^2 - 3*3) - (1^2 - 3*1)#

#= (9-9) - (1-3) = 0 -(-2)#

#= 2#

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Answer 2

Scarlett Allison

To evaluate the definite integral of ( \int_{1}^{3} (2x - 3) , dx ), you first integrate the function ( 2x - 3 ) with respect to ( x ) and then evaluate it at the upper and lower limits of integration (1 and 3, respectively).

The antiderivative of ( 2x - 3 ) with respect to ( x ) is ( x^2 - 3x ).

Substituting the upper and lower limits of integration into this antiderivative gives:

[ F(3) - F(1) = (3^2 - 3 \times 3) - (1^2 - 3 \times 1) = (9 - 9) - (1 - 3) = 0 - (-2) = 2 ]

So, the value of the definite integral ( \int_{1}^{3} (2x - 3) , dx ) is 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.