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

Question

The values of a function $f(x)$ at different values of $x$ are as follows

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

Aicharya · Accepted Answer

$h=$ difference of two values of $x$ Take value of $f(x)$ as $\left(y_0, y_1, y_2, \ldots, y_5\right)$ Then by Trapezoidal rule Now, $\int_{x_0}^{x_0+n h} f(x) d x$ $=\frac{h}{2}\left[\left(y_0+y_5\right)+2\left(y_1+y_2+y_3+y_4\right)\right]$ $=\frac{1}{2}[(2+27)+2(3+6+11+18)]$ $=\frac{1}{2}[29+2 \cdot 38]=\frac{1}{2}(29+76)$ $=\frac{1}{2} \times 105=52.5$ Approximate area $=52.5 \mathrm{sq}$ unit

The values of a function $f(x)$ at different values of $x$ are as follows Then, the approximate area (in…

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