The values of a function f(x) at different va | vedclass

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(C) Given $h = 1$ (the difference between consecutive $x$ values).Let the values of $f(x)$ be $y_0, y_1, y_2, y_3, y_4, y_5$.Here, $y_0 = 2, y_1 = 3,

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$52.5$

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(C) Given $h = 1$ (the difference between consecutive $x$ values).Let the values of $f(x)$ be $y_0, y_1, y_2, y_3, y_4, y_5$.Here, $y_0 = 2, y_1 = 3, y_2 = 6, y_3 = 11, y_4 = 18, y_5 = 27$.By the Trapezoidal rule, the area $A$ is given by:$A = \frac{h}{2} [ (y_0 + y_5) + 2(y_1 + y_2 + y_3 + y_4) ]$Substituting the values:$A = \frac{1}{2} [ (2 + 27) + 2(3 + 6 + 11 + 18) ]$$A = \frac{1}{2} [ 29 + 2(38) ]$$A = \frac{1}{2} [ 29 + 76 ]$$A = \frac{1}{2} [ 105 ] = 52.5$Thus, the approximate area is $52.5$ square units.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

The values of a function $f(x)$ at different values of $x$ are as follows:
$x$ $0$ $1$ $2$ $3$ $4$ $5$
$f(x)$ $2$ $3$ $6$ $11$ $18$ $27$

Then, the approximate area (in square units) bounded by the curve $y=f(x)$ and the $x$-axis between $x=0$ and $x=5$, using the Trapezoidal rule, is:

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The values of a function $f(x)$ at different values of $x$ are as follows:$x$$0$$1$$2$$3$$4$$5$$f(x)$$2$$3$$6$$11$$18$$27$Then, the approximate area (in square units) bounded by the curve $y=f(x)$ and the $x$-axis between $x=0$ and $x=5$, using the Trapezoidal rule, is: