#### To determine

**To find:**

A function fx and the value of the constant a

#### Answer

fx=cosx and a=π6

#### Explanation

**1) Concept:**

i) The First Fundamental Theorem for calculus: If f is continuous on [a, b], then the function g is defined by g(x)=∫abftdta≤x≤b is continuous on [a,b] and differentiable on (a,b) and g’(x)=f(x)

ii) The Second Fundamental Theorem for calculus: If f is continuous on [a, b], then the function is defined by ∫abftdt=Fb-Fa𝑤here F is an antiderivative of f

**2) Given:**

2∫axftdt=2sinx-1

**3) Calculation:**

The given integral is

2∫axftdt=2sinx-1

∫axftdt=2sinx-12

Assuming f is continuous let us apply first fundamental theorem of calculus. Therefore take derivative of both sides with respect to x

ddx∫axftdt=ddx(sinx-12)

Here all the terms are a function of x

Therefore the equation becomes

fx=cosx

Now to find the value of a, use the second fundamental theorem

∫abftdt=Fb-Fa

∫abftdt =∫axcost dt=sintax

=sinx-sina

But the value of integral is given as 2sinx-12

Therefore compare both values

2sinx-12=sinx-sina

2sinx-1=2sinx-2sina

-1=-2sina

sina=12

a=sin-112

a=π6

**Conclusion:**

Therefore, fx=cosx and a=π6