how is this solved?

Question

1. Find the average value of the function over the given interval.
(a) h(x) = 2x + 2 over [1, 3]
(b) f(x) = e2x over [0, 10]

Answer 1

Like this:

Integrate the function, find the difference of the integrand between the two limits, and divide that by the length of the range. So:

(a) Integrand is [x^2 + 2 x]; difference between limits is:

(3^2 - 1^2) + 2 (3 - 1) = 8 + 4 = 12. Range length = 2, so:

For (a), the AVERAGE VALUE = 6.
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(b) Integrand is (1/2) e^(2x) *** ; difference between limits is:

[(1/2) e^(2x)]. Range length = 10, so:

For (b), the AVERAGE VALUE = (1/20) [e^20 - 1], or 24258259.72..., for what that's worth.

(Note: the "- 1" only affects the last significant figure here! The decimal part would be .77 without it.)
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*** By the way, please try to write exponents in the manner generally used by people in this forum or service, and indeed by people using computers to write mathematics with ordinary text or with sophisticated tools like Latek : e^(2x). There are TWO things to note here which benefit clear communication:

(i) The "^" ("carat") symbol SIGNALS that what follows is AN EXPONENT. You can think of it as saying "up there".

(ii) Use PARENTHESES for clarity. Writing e^2 is unambiguous; however, e^2x wouldn't be, just like your own e2x. (Would e^2x mean "e to the 2x." or "e squared times x"? Could e2x mean "e to the 2x" or "e squared times x" or "e times 2 times x" --- a triple ambiguity?) The parentheses around an expression clearly delineate something that should be thought of as ONE ENTITY, once evaluated.

Live long and prosper.