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

The correct option is 1.

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)=(d)/(dx)f(x)`

`f'(x)=(d)/(dx)(\int_0^x\sin (t^2)dt)`

`f'(x)=\sin (x^2)`
`[\because (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.

Answer 2

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.