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Assume the formula \sum \limits ^n_{i=1}i^5= \frac {2n^6+6n^5+5n^4-n^2}{12} holds for every n \in...

Question:

Assume the formula {eq}\sum \limits ^n_{i=1 }i^5= \frac {2n^6+6n^5+5n^4-n^2}{12} {/eq} holds for every {eq}n \in \mathbb{N} {/eq}. Fix {eq}b>0. {/eq}

Consider the partition {eq}0=x_0<x_1<x_2<...<x_{n-1}<x_n=b {/eq} of the interval {eq}[0,b] {/eq} into {eq}n {/eq} equal parts, and let {eq}x^*_i=x_i {/eq} be the sample point in the interval {eq}[x_{i-1},x_i] {/eq}

Use the definition of the definite integral, to compute {eq}\int_0^b x^5 dx {/eq}

Evaluating a Definite Integral


The question asks to evaluate a definite integral of a function. Using the Fundamental Theorem of Calculus which is used in Calculus to evaluate definite integrals, we calculate the value of the definite integral.


Answer and Explanation:


The anti-derivative of the function {eq}f(x)=x^5 {/eq} from Calculus is given by

{eq}\displaystyle F(x) = \frac {x^{5+1}}{5+1} = \frac {x^6}{6} \qquad (1) {/eq}

Hence from the Fundamental Theorem of Calculus

{eq}\displaystyle \int_0^b x^5 \; dx = F(b)-F(0) = \frac {b^6}{6}-\frac {0^6}{6} = \frac {b^6}{6}. {/eq}

The answer is {eq}\displaystyle \frac {b^6}{6}. {/eq}


Learn more about this topic:

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Evaluating Definite Integrals Using the Fundamental Theorem

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2
1.8K

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