#### To determine

**To explain:**

What is wrong in the given equation.

#### Answer

The given expression is incorrect, since ∫0πsec2x dx does not exist as a difference of anti-derivatives.

#### Explanation

**1) Concept:**

Use the fundamental theorem of calculus part 2 to explain the wrong in the given equation.

**2) Fundamental theorem of Calculus, Part **2

If f is continuous on a,b, then

∫abfxdx=Fb-F(a)

where F is any antiderivative of f, that is, a function F such that F'=f

**2) Given:**

∫0πsec2x dx=tanx]π0=0

**3) Calculation:**

The graph of function fx=sec2x, 0≤x≤π is given by,

The function fx=sec2x is not continuous on 0,π

So, the Fundamental theorem of Calculus, Part 2 cannot be applied. So the expression

fx=sec2x=Fπ-F0 where F is antiderivative of f, that is, a function F such that F'(x)=f(x) is wrong.

**Conclusion:**

∫0πsec2x dx does not exists as a difference of anti-derivatives, therefore the given statement is wrong.