Question

F of x equals the integral from 0 to x of the sine of t squared . use your calculator to find f '(1). 0.841 0.709 0.292 0.540

Answer 1

Analytically, the result will be ...

... f'(1) = sin(1)² ≈ 0.70807341827357...

My calculator shows the same result.

The most appropriate choice appears to be 0.709.

Answer 2

Answer:

The correct option is 1.

Step-by-step explanation:

The given function is

$f(x)=\int_0^x\sin (t^2)dt$

We need to find the value of f '(1).

$f'(x)=\frac{d}{dx}f(x)$

$f'(x)=\frac{d}{dx}(\int_0^x\sin (t^2)dt)$

$f'(x)=\sin (x^2)$          $[\because \frac{d}{dx} (\int_0^xf(t)dt)=f(x)]$

Substitute x=1 in the above function to find the value of f'(1).

$f'(1)=\sin (1^2)$

$f'(1)=0.841470984808$

$f'(1)=0.841$

The value of f'(1) is 0.841. Therefore the correct option is 1.