Evaluate the integral.

{eq}\displaystyle \int_0^{\dfrac \pi 6} \sqrt {1 + \cos 2 x}\ dx {/eq}


Evaluate the integral.

{eq}\displaystyle \int_0^{\dfrac \pi 6} \sqrt {1 + \cos 2 x}\ dx {/eq}

Fundamental Theorem of Calculus:

For evaluating the final result of the definite integral, the theorem that is used is known as the Fundamental theorem of calculus, part 2. This theorem states that to solve a definite integral first we have to find the antiderivative of the function and then evaluate the value of the antiderivative at the boundaries of the given integral and then subtract these two values.

$$\int_{a}^{b}f(x)dx=F(b)-F(a) $$

{eq}F(x) {/eq} is an antiderivative of {eq}f(x) {/eq}

Answer and Explanation: 1

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$$\displaystyle \int_0^{\dfrac{\pi}{6}} \sqrt{1 + \cos 2 x}\ dx $$

We have to evaluate the integral.

The given definite integral can be...

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Learn more about this topic:

Evaluating Definite Integrals Using the Fundamental Theorem


Chapter 16 / Lesson 2

In calculus, the fundamental theorem is an essential tool that helps explain the relationship between integration and differentiation. Learn about evaluating definite integrals using the fundamental theorem, and work examples to gain understanding.

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