# Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 276: 72

$$\frac{38}{3}$$

#### Work Step by Step

Given $$\int_{1}^{6} \sqrt{x+3} d x$$ Let $$u= x+3 \ \ \ \ \Rightarrow \ \ \ du = dx$$ At $x=1 \to u= 4$ and at $x=6 \to u=9$ Then \begin{aligned} \int_{1}^{6} \sqrt{x+3} d x &=\int_{4}^{9} \sqrt{u} d u \\ &=\left.\frac{2}{3} u^{3 / 2}\right|_{4} ^{9} \\ &=\frac{2}{3}\left(9^{3 / 2}-4^{3 / 2}\right) \\ &=\frac{2}{3}(27-8)=\frac{2}{3}(19)\\ &=\frac{38}{3} \end{aligned}

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