## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 461: 100

#### Answer

$$25.976$$

#### Work Step by Step

Given $$\int_{2}^{4} \sqrt{6 t^{3}+1} d t$$ Since $\Delta x=\dfrac{b-a}{n}=\dfrac{2}{3}$, then \begin{align*} T_{n}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+..+2f(x_{n-1})+f(x_n)\right]\Delta x\\ T_{3}&= \dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+2f(x_{3}) \right]\Delta x\\ &=\dfrac{1}{3}(\sqrt{6(2)^{3}+1}+2 \sqrt{6\left(\frac{8}{3}\right)^{3}+1}+2 \sqrt{6\left(\frac{10}{3}\right)^{3}+1}+\sqrt{6(4)^{3}+1})\\ &\approx 25.976 \end{align*}

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