Compute h → 0 lim h 1 ∫ 1 1 + 2 h e x sin ( 3 π x ) d x .
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@Mukhammadsaid Jr. , you can enter as \displaystyle \lim {h\to 0} \frac 1h \int 1^{1+2h} e^{\sqrt x}\sin \left(\frac{\pi x}3 \right) dx h → 0 lim h 1 ∫ 1 1 + 2 h e x sin ( 3 π x ) d x . Note that a lot of { } are unnecessary. \left and \right automatically adjust the brackets size. If you are using \ [ \ ] instead of \ ( \ ) , the \displaystyle is not necessary but the formula will be centralized as follows:
h → 0 lim h 1 ∫ 1 1 + 2 h e x sin ( 3 π x ) d x
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Let f ( x ) = e x sin ( 3 π x ) d x and F ( x ) = ∫ 0 x f ( t ) d t . Then we have:
L = h → 0 lim h 1 ∫ 1 1 + 2 h e x sin ( 3 π x ) d x = h → 0 lim h F ( 1 + 2 h ) − F ( 1 ) = 2 h → 0 lim 2 h F ( 1 + 2 h ) − F ( 1 ) = 2 f ( 1 ) = 2 e sin 3 π ≈ 4 . 7 0 8 By definition of differentiation