How do you evaluate #int 2x^37# for [2,5]?
This is basically finding f(x) with the diffrential given.
So integration, (and points)
I'll just ignore typing the integral sign.
So first split the equation. Int. of 2x^3  Int. of 7 power of x increases +1 and becomes denominator, so.
x^4/2  7x
y=x^4/2  7x + c
Insert 2,5 5=(2)^4/2  7(2) + c 5= 6+c 11=c
=> y=(x^4)/27x+11
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To use the Fundamental Theorem of Calculus.
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To evaluate the expression 2x^3  7 for the interval [2, 5], you need to substitute each endpoint of the interval into the expression and calculate the result.
When x = 2: 2(2)^3  7 = 2(8)  7 = 16  7 = 9
When x = 5: 2(5)^3  7 = 2(125)  7 = 250  7 = 243
So, the values of the expression 2x^3  7 for the interval [2, 5] are 9 and 243, respectively.
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To evaluate ( \int_{2}^{5} (2x^3  7) , dx ), you need to find the antiderivative of ( 2x^3  7 ) with respect to ( x ), and then evaluate the antiderivative at the upper limit of integration (5) and subtract the value of the antiderivative at the lower limit of integration (2).

Find the antiderivative of ( 2x^3  7 ): [ \int (2x^3  7) , dx = \frac{1}{2}x^4  7x + C ]

Evaluate the antiderivative at the upper limit of integration: [ \left[ \frac{1}{2}x^4  7x \right]_{2}^{5} ] [ = \left( \frac{1}{2}(5)^4  7(5) \right)  \left( \frac{1}{2}(2)^4  7(2) \right) ]

Calculate the result: [ = \left( \frac{1}{2}(625)  35 \right)  \left( \frac{1}{2}(16)  14 \right) ] [ = \left( 312.5  35 \right)  \left( 8  14 \right) ] [ = (277.5)  (6) ] [ = 283.5 ]
So, ( \int_{2}^{5} (2x^3  7) , dx = 283.5 ).
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When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
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