Question

Find the derivative of the function \(y = \int_0^{\tan x} {\sqrt {t + \sqrt t } } dt\) using Part 1 of The Fundamental Theorem of Calculus.

Short answer

The derivative of function \(y = \int_0^{\tan x} {\sqrt {t + \sqrt t } } dt\) is \(\sqrt {\tan x + \sqrt {\tan x} } \).

Step-by-step solution

The Fundamental Theorem of Calculus, Part 1.

If\(f\)is continuous on\({\rm{(a, b)}}\), then the function\({\rm{g}}\)is defined by\(g(x) = \int_a^x f (t)dt,a \le x \le b\)is an anti-derivative of\(f\), that is,\({g^\prime }(x) = f(x)\)for\({\rm{a < x < b}}\).

 Step 2: Use The Fundamental Theorem of Calculus for calculation.

The given function is \(y = \int_0^{\tan x} {\sqrt {t + \sqrt t } } dt\).

Obtain the derivative of the function \(y = \int_0^{\tan x} {\sqrt {t + \sqrt t } } dt\), using the above-mentioned theorem as follows.

As per the fundamental theorem of calculus, if for \(a\) continuous function \(f\), \(g(x) = \int_a^x f (t)dt\), then \({g^\prime }(x) = f(x)\).

Therefore, by comparing the given function with the theorem, it is obtained that, \({\rm{g(x) = y}}\) and \(f(t) = \sqrt {t + \sqrt t } \). Then, \({y^\prime }(x) = f(x)\).

Therefore, by applying the above-mentioned theorem, the derivative of \(y\)is obtained as \({y^\prime }(x) = \sqrt {\tan x + \sqrt {\tan x} } \).

Therefore, the derivative of function \(y = \int_0^{\tan x} {\sqrt {t + \sqrt t } } dt\) is \(\sqrt {\tan x + \sqrt {\tan x} } \).