How do you find the value of c guaranteed by the mean value theorem for integrals #2sec^2x#?

Answer 1

An interval needs to be specified here, before proceeding any further.

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Answer 2

To find the value of ( c ) guaranteed by the Mean Value Theorem for integrals for the function ( f(x) = 2\sec^2(x) ) over the interval ([a, b]):

  1. Compute the definite integral of ( f(x) ) over the interval ([a, b]) using any suitable method. Let this integral be denoted by ( F(b) - F(a) ), where ( F(x) ) is the antiderivative of ( f(x) ).

  2. Apply the Mean Value Theorem for Integrals, which states that there exists a ( c ) in the interval ([a, b]) such that: [ \int_a^b f(x) , dx = f(c) \times (b - a) ]

  3. Set up the equation using the values obtained: [ \int_a^b 2\sec^2(x) , dx = 2\sec^2(c) \times (b - a) ]

  4. Solve for ( c ) using the equation obtained in step 3.

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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.

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.

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.

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.

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