Question

(a) integrate to find \(F\) as a function of \(x\) and (b) demonstrate the Second Fundamental Theorem of Calculus by differentiating the result in part (a). $$ F(x)=\int_{\pi / 4}^{x} \sec ^{2} t d t $$

Short answer

The function after integration is \(F(x) = \tan x - 1\). The derivative of this function (which also demonstrates the Second Fundamental Theorem of Calculus) is \(F'(x) = \sec^2 x\).

Step-by-step solution

Carry out the Integration of \(F(x)\)

First, we find the integral of \(F(x)\) which is given as \(F(x) = \int_{\pi/4}^x \sec^2 t\, dt\) . Knowing that the integral of \(\sec^2 t\) is \(\tan t\), we substitute the limits \(\pi/4\) and \(x\). Therefore, we have \(F(x) = [\tan t]_{\pi/4}^x = \tan x - \tan (\pi/4)\). Since \(\tan (\pi/4) = 1\), we simplify to obtain \(F(x) = \tan x - 1\).

Apply the Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus states that if a function is a continuous real-valued function on an interval \([a, b]\) and its integral from a to any point \(x\) in the interval is given by \(F(x)\), then the derivative of \(F(x)\) is \(f(x)\). Therefore, we find the derivative of \(F(x) = \tan x - 1\). This gives us \(F'(x) = \sec^2 x\) as the derivative of \(\tan x\) is equal to \(\sec^2 x\). The derivative of \(1\) is zero, so we don't include it in the final result.