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f ' (1) = sin(1) * cos(1).

You can evaluate f(1) until you know the function f(t).

I'll first show you two possible forms of the function f(t). Then I'll review the technique (integration by substitution) that would typically be used to integrate sin t * cos t.

The general antiderivative of sin t * cos t is either

1/2 sin^2(t) + c

or

-1/2 cos^2(t) + c,

where c in either case is an arbitrary constant.

Before reading further, if you haven't already done so you can (and should) verify that the derivative of either form gives you sin t * cos t.

The two forms look different, but since

sin^2(t) + cos^2(t) = 1

it follows, for example, that

First recall that

sin^2(t) = 1 - cos^2(t).

From this we see that

1/2 sin^2(t) + c

can be written as

1/2 (1 - cos^2(t)) + c

= -1/2 cos^2(t) + 1/2 + c.

Recall that c for either form is an arbitrary constant, meaning that c can take any value.

Since c can take any value, 1/2 + c can also take any value. Thus 1/2 + c (which can take any value) can just be expressed as c (which can take any value). This property of arbitrary constants is often expressed by saying that the arbitrary constant c can 'absorb' the 1/2, or for that matter any other constant number.

Thus our original

1/2 sin^2(t) + c

can be written as

-1/2 cos^2(t) + c,

where we repeat that c in either case is to be regarded as an arbitrary constant.

Now the next thing we need to do to solve this problem is pick one of the two forms, and find the value of c that goes with that form.

Picking the first form

f(t) = 1/2 sin^2(t) + c

we apply the condition that f(0) = 1. Since f(0) = 1/2 sin^2(0) + c = c, we find that the condition

f(0) = 1

becomes

c = 1.

So f(t) = 1/2 sin^2(t) + 1 solves the given equation with the given condition.

Evaluating f(b) for b = 0, 1/2, 1, 3/2, 2, 5/2, 3 yields the requested solution.

Had we used the form

f(t) = -1/2 cos^2(t) + c

we would get f(0) = 1/2 cos^2(0) + c = -1/2 * 1 + c = -1/2 + c. Thus our initial condition

f(0) = 1

would become

-1/2 + c = 1

so that c = 3/2. Our function would therefore be

f(t) = -1/2 cos^2(t) + 3/2.

If you evaluate f(b) for the various given values of b you will get the same results you obtained using f(t) = 1/2 sin^2(t) + 1.

The reason you get the same results is straightforward enough:

1/2 sin^2(t) + 1 = 1/2 ( 1 - cos^2(t) ) + 1 = 1/2 - cos^2(t) + 1 = -cos^2(t) + 3/2.

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I don't know the context of this question, so I don't know how you were expected to know how to find an antiderivative of sin t * cos t. The technique of integration by substitution isn't covered until Chapter 7.

However I'll apply the technique here, and if you haven't yet gotten to chapter 7 you can refer back to this once you have covered the technique.

To integrate sin t * cos t with respect to t, we use a substitution (sometimes also called a change of variable).

In this case our integrand can be written

(sin t * cos t) dt

If can let

u = sin t

then

du / dt = cos t

so

du = (cos t) dt.

Since

(sin t * cos t) dt

can be written

sin t ( (cos t ) dt)

and since u = sin t and du = (cos t) dt this becomes

u * du.

The integral of u du is just the integral of u with respect to u. Thus

integral ( u du ) = u^2 / 2 + c.

Now since u = sin(t) we see that our original integral must be

(sin t)^2 / 2 + c.

We could alternatively chosen to let

u = cos(t)

so that

du/dt = -sin(t)

and

du = -sin t dt.

The our original integrand

(sin t * cos t) dt

would become

- u du

and our integral would be

integral(-u du) = - u^2 / 2 + c.

Since u = cos t this is just

-(cos t)^2 / 2 + c.

Once more this part will probably make more sense after a couple more assignments, but once you understand the context it fits in nicely with the preceding note.

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