Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
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
Solution:
2336 Upvotes
Verified Answer
The correct answer is:
52.5
$h=$ 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
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.