Find f'(0.3) for f(x)= the integral from 0 to x of arccos(t)dt

1 answer:

8 0

$\bf \textit{using the 2nd fundamental theorem of calculus}\\\\
\cfrac{dy}{dx}\displaystyle \left[ \int\limits_{0}^{x}\ cos^{-1}(t)dt \right]\implies cos^{-1}(x)
\\\\\\
f'(0.3)\iff cos^{-1}(0.3)\approx 1.26610367277949911126$

now.. 0.3 is just a value...we'e assuming Radians for the inverse cosine, so, if you check, make sure your calculator is in Radian mode

You might be interested in

Which of the following represents the graph of this equation? y = 1/2|x|

Aliun [14]

The graph is shown in the attached image.

6 0

2 years ago

(a) Use the definition to find an expression for the area under the curve y = x3 from 0 to 1 as a limit. lim n→∞ n i = 1 Correct

Luba_88 [7]

Splitting up the interval of integration into $n$ subintervals gives the partition

$\left[0,\dfrac1n\right],\left[\dfrac1n,\dfrac2n\right],\ldots,\left[\dfrac{n-1}n,1\right]$

Each subinterval has length $\dfrac{1-0}n=\dfrac1n$ . The right endpoints of each subinterval follow the sequence

$r_i=\dfrac in$

with $i=1,2,3,\ldots,n$ . Then the left-endpoint Riemann sum that approximates the definite integral is

$\displaystyle\sum_{i=1}^n\frac{{r_i}^3}n$

and taking the limit as $n\to\infty$ gives the area exactly. We have

$\displaystyle\lim_{n\to\infty}\frac1n\sum_{i=1}^n\left(\frac in\right)^3=\lim_{n\to\infty}\frac{n^2(n+1)^2}{4n^3}=\boxed{\frac14}$

6 0

3 years ago

What is the lowest common multiple (LCM) of 3 and 4

Kazeer [188]

Answer:

Step-by-step explanation:

The Least Common Multiple (LCM) of some numbers is the smallest number that the numbers are factors of. Like the LCM of 3 and 4 is 12, because 12 is the smallest number that 3 and 4 are both factors for.

8 0

3 years ago

Read 2 more answers

Can you please help me? I forgot this stuff. . .

Lelu [443]

WHHEEEEEZZZZZZZ

I HAVE AESTHEMA

5 0

3 years ago

Read 2 more answers

Can you please help me?

zubka84 [21]