((a^2m+n)(b^n-2m))/((a^n-m)(b^2m-n))

Mathematics

1 answer:

Novosadov [1.4K]2 years ago

7 0

Answer:

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Step-by-step explanation:

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Natalija [7]

Split up the interval [2, 5] into $n$ equally spaced subintervals, then consider the value of $f(x)$ at the right endpoint of each subinterval.

The length of the interval is $5-2=3$ , so the length of each subinterval would be $\dfrac3n$ . This means the first rectangle's height would be taken to be $x^2$ when $x=2+\dfrac3n$ , so that the height is $\left(2+\dfrac3n\right)^2$ , and its base would have length $\dfrac{3k}n$ . So the area under $x^2$ over the first subinterval is $\left(2+\dfrac3n\right)^2\dfrac3n$ .

Continuing in this fashion, the area under $x^2$ over the $k$ th subinterval is approximated by $\left(2+\dfrac{3k}n\right)^2\dfrac{3k}n$ , and so the Riemann approximation to the definite integral is

$\displaystyle\sum_{k=1}^n\left(2+\frac{3k}n\right)^2\frac{3k}n$

and its value is given exactly by taking $n\to\infty$ . So the answer is D (and the value of the integral is exactly 39).