6EDefiniteIntegrals

Definite Integrals

Review theory of integral calculus

Example 1

math . \quad \displaystyle{ \int \limits_{-1}^{1} } \dfrac{1} {\sqrt{4-x^2}} \, dx math . math . \qquad \qquad \qquad \; = \left[ \text{Sin}^{-1} \dfrac{x}{2} \right]_{-1}^{1} math . math . \qquad \qquad \qquad\; = \text{Sin}^{-1} \left( \dfrac{1}{2} \right) - \text{Sin}^{-1} \left( \dfrac{-1}{2} \right) math . math \\ . \qquad \qquad \qquad\; = \left( \dfrac{\pi}{6} \right) - \left( \dfrac{-\pi}{6} \right) \\. \\ . \qquad \qquad \qquad\; = \dfrac{2\pi}{6} \\. \\ . \qquad \qquad \qquad\; = \dfrac{\pi}{3} \; \text{ square units} math

Example 2

math . \quad \displaystyle{\int\limits_{x=0}^{x=1} } \dfrac{3x}{\sqrt{x+1}} \, dx math

Include variable (x) in terminals to help keep track of changes to variable during substitution method.

math \text{Let } \; u=x+1 \; \text{ so } \; \dfrac{du}{dx}=1 \; \text{ so } \; dx=du math . math \\ \text{And when } \; x=0, \; u=1 \\ \text{And when } \; x=1, \; u=2 math . math . \quad \displaystyle{\int\limits_{x=0}^{x=1} } \dfrac{3x}{\sqrt{x+1}} \, dx \; = \displaystyle{\int\limits_{u=1}^{u=2} } \dfrac{3(u-1)}{\sqrt{u}} \, du math . Notice change of variable to u in the terminals

math . \qquad \qquad \qquad \; = \displaystyle{ 3\int\limits_{u=1}^{u=2} (u-1)u^{-\frac{1}{2}} \; du} math . math . \qquad \qquad \qquad \; = \displaystyle{ 3\int\limits_{u=1}^{u=2} u^{\frac{1}{2}}-u^{-\frac{1}{2}} \; du } math . math . \qquad \qquad \qquad \; = 3\left[\dfrac{2}{3} u^{\frac{3}{2}}-2u^{\frac{1}{2}} \right]_{u=1}^{u=2} math . math . \qquad \qquad \qquad \; = 3\left[ \left( \dfrac{2}{3}(2)^{\frac{3}{2}}-2(2)^{\frac{1}{2}} \right) - \left( \dfrac{2}{3}(1)^{\frac{3}{2}}-2(1)^{\frac{1}{2}} \right) \right] math . math . \qquad \qquad \qquad \; = 3\left[ \left( \dfrac{2^{\frac{5}{2}}}{3}-2^{\frac{3}{2}} \right) - \left( \dfrac{2}{3}-2 \right) \right] math . math . \qquad \qquad \qquad \; = 3\left[ \left( \dfrac{2^{\frac{5}{2}}}{3}-\dfrac{3\times 2^{\frac{3}{2}}}{3} \right) - \left( -\dfrac{4}{3} \right) \right] math . math . \qquad \qquad \qquad \; = 2^{\frac{5}{2}}-3\times 2^{\frac{3}{2}} + 4 math . math \\ . \qquad \qquad \qquad \; = 4\sqrt{2}-3\times 2\sqrt{2} + 4 \\. \\ . \qquad \qquad \qquad \; = -2\sqrt{2} + 4 \; \text{square units} math

Definite Integrals on the Calculator

Recall that your calculator can perform indefinite integrals. It can also perform definite integrals using the same form. In the MAIN screen, bring up the virtual keyboard. In the **2D** keyboard, select **CALC** from the bottom row and select the Integral form (see screen capture below)

Example 1 Make sure you have your calculator mode set to **REAL**, **RADIANS**, **STANDARD**

Enter the integral math \displaystyle{\int\limits_{-1}^{1} } \dfrac{1}{\sqrt{4-x^2}}dx math Don't forget to put the x in the end of the form to make **dx**



Example 2 Enter the integral math \displaystyle{\int\limits_{0}^{1} } \dfrac{3x}{\sqrt{x+1}}dx math



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