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**Section 5.3 – The Definite Integral**

As the number of rectangles increased, the approximation of the area under the curve approaches a value. Copyright 2010 Pearson Education, Inc.

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**Section 5.3 – The Definite Integral**

Definition Note: The function f(x) must be continuous on the interval [a, b].

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**Section 5.3 – The Definite Integral**

Parts of the Definite Integral Copyright 2010 Pearson Education, Inc.

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**Section 5.3 – The Definite Integral**

Properties of the Definite Integral Copyright 2010 Pearson Education, Inc.

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**Section 5.3 – The Definite Integral**

Geometric Interpretations of the Properties of the Definite Integral Copyright 2010 Pearson Education, Inc.

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**Section 5.3 – The Definite Integral**

Using the Properties of the Definite Integral 1 3 𝑓 𝑥 𝑑𝑥 = 𝑓 𝑥 𝑑𝑥 = 𝑔 𝑥 𝑑𝑥 =−4 Given: 1 3 3𝑓 𝑥 𝑑𝑥 = 3 1 3 𝑓 𝑥 𝑑𝑥 = 3 6 =18 1 3 2𝑓 𝑥 −4𝑔 𝑥 𝑑𝑥 = 2 1 3 𝑓 𝑥 𝑑𝑥 −4 1 3 𝑔 𝑥 𝑑𝑥 = 2 6 −4 −4 =28 1 7 𝑓 𝑥 𝑑𝑥 = 1 3 𝑓 𝑥 𝑑𝑥 𝑓 𝑥 𝑑𝑥 = 6+9=15 3 1 𝑓 𝑥 𝑑𝑥 = − 1 3 𝑓 𝑥 𝑑𝑥 = −6

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**Section 5.3 – The Definite Integral**

Rules of the Definite Integral 𝑎 𝑏 𝑐 𝑑𝑥=𝑐(𝑏−𝑎) 𝑎 𝑏 𝑥 𝑑𝑥= 𝑏 2 2 − 𝑎 2 2 𝑎 𝑏 𝑥 2 𝑑𝑥= 𝑏 3 3 − 𝑎 3 3 Examples 2 6 4 𝑑𝑥= 4 8 𝑥 𝑑𝑥= − = 4 6−2 = 16 32−8= 24 3 5 𝑥 2 𝑑𝑥 = − = 125 3 − 27 3 = 98 3 =32.67 3 4 𝑥 2 +3𝑥−2 𝑑𝑥 = 3 4 𝑥 2 𝑑𝑥 𝑥 𝑑𝑥 − 𝑑𝑥 = − − − 2 4−3 = − 9 2 − 64 3 − 2 1 = 20.83

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**Section 5.4 – The Fundamental Theorem of Calculus**

𝐼𝑓 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑒𝑣𝑒𝑟𝑦 𝑝𝑜𝑖𝑛𝑡 𝑖𝑛 𝑎, 𝑏 𝑎𝑛𝑑 𝐹 𝑖𝑠 𝑎𝑛𝑦 𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑓 𝑜𝑛 𝑎, 𝑏 , 𝑡ℎ𝑒𝑛 𝑎 𝑏 𝑓 𝑥 𝑑𝑥=𝐹 𝑏 −𝐹(𝑎) . Examples 1 5 5𝑥 𝑑𝑥 = 5 𝑥 = 5 1 5(5) 2 2 − = 125 2 − 5 2 = 120 2 =60 𝜋 6 2𝜋 3 𝑠𝑖𝑛𝑥 𝑑𝑥 = 5 1 −𝑐𝑜𝑠 2𝜋 3 − −𝑐𝑜𝑠 𝜋 6 = − − 1 2 − − = −𝑐𝑜𝑠𝑥 = 0.866

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**Section 5.4 – The Fundamental Theorem of Calculus**

𝑎 𝑏 𝑓 𝑥 𝑑𝑥=𝐹 𝑏 −𝐹(𝑎) . Examples 3 4 𝑥 2 +3𝑥−2 𝑑𝑥 = 𝑥 𝑥 2 2 −2𝑥 = 4 3 (4) 2 2 −2(4) − (3) 2 2 −2(3) 37.33−16.5= 20.83 𝑥 −1 5 − = 32 1 𝑥 𝑑𝑥 = 32 1 32 𝑥 −6 5 𝑑𝑥 = −5 𝑥 = − 5 2 − − 5 1 = 2.5 1

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**Section 5.4 – The Fundamental Theorem of Calculus**

Differentiating a Definite Integral 𝐼𝑓 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑜𝑛 𝑎, 𝑏 𝑎𝑛𝑑 𝑥 𝑖𝑠 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑎, 𝑏 , 𝑡ℎ𝑒𝑛 𝐹 ′ 𝑥 = 𝑑 𝑑𝑥 𝑎 𝑥 𝑓 𝑡 𝑑𝑡=𝑓(𝑥) 𝑑 𝑑𝑥 1 𝑥 𝑡 2 𝑑𝑡= 𝑑 𝑑𝑥 𝑡 = 𝑥 1 𝑑 𝑑𝑥 𝑥 3 3 − = 𝑑 𝑑𝑥 𝑥 3 3 − 1 3 = 𝑥 2 𝑑 𝑑𝑥 1 4𝑥 𝑡 2 𝑑𝑡= 𝑑 𝑑𝑥 𝑡 = 4𝑥 1 𝑑 𝑑𝑥 (4𝑥) 3 3 − = 𝑑 𝑑𝑥 64𝑥 3 3 − 1 3 = 64𝑥 2 𝑑 𝑑𝑥 1 𝑥 2 𝑡 2 𝑑𝑡= 𝑑 𝑑𝑥 𝑡 = 𝑥 2 1 𝑑 𝑑𝑥 ( 𝑥 2 ) 3 3 − = 𝑑 𝑑𝑥 𝑥 6 3 − 1 3 = 6 𝑥 5 3 = 2𝑥 5

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**Section 5.4 – The Fundamental Theorem of Calculus**

Differentiating a Definite Integral 𝐼𝑓 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑜𝑛 𝑎, 𝑏 𝑎𝑛𝑑 𝑥 𝑖𝑠 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑎, 𝑏 , 𝑡ℎ𝑒𝑛 𝐹 ′ 𝑥 = 𝑑 𝑑𝑥 𝑎 𝑥 𝑓 𝑡 𝑑𝑡=𝑓(𝑥) 𝑑 𝑑𝑥 1 𝑥 𝑡 2 𝑑𝑡= 𝑑 𝑑𝑥 1 𝑥 2 sin 𝑡 𝑑𝑡= 𝑥 = 𝑥 2 2𝑥 𝑠𝑖𝑛 𝑥 2 𝑑 𝑑𝑥 1 4𝑥 𝑡 2 𝑑𝑡= 𝑑 𝑑𝑥 1 𝑥 2 +𝑥 tan 𝑡 𝑑𝑡= 4𝑥 = 64𝑥 2 2𝑥+1 𝑡𝑎𝑛 𝑥 2 +2 𝑑 𝑑𝑥 1 𝑥 2 𝑡 2 𝑑𝑡= 𝑑 𝑑𝑥 𝑡𝑎𝑛𝑥 𝑠𝑖𝑛𝑥 𝑡 2 𝑑𝑡= 𝑥 𝑥 = 2𝑥 5 (𝑠𝑖𝑛 2 𝑥) 𝑐𝑜𝑠𝑥 −(𝑡𝑎𝑛 2 𝑥) 𝑠𝑒𝑐 2 𝑥

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**Section 5.4 – The Fundamental Theorem of Calculus**

Mean Value Theorem for Integrals Copyright 2010 Pearson Education, Inc.

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**Section 5.3 – The Definite Integral**

Mean Value for Definite Integral Find the mean (average) value of the function 𝑓 𝑥 = 𝑥 2 +2 over the interval 1, 3 𝐴𝑉= 1 3− 𝑥 𝑑𝑥 3 𝐴𝑉= 𝑥 𝑥 1 𝐴𝑉= − 𝐴𝑉= − 7 3 𝐴𝑉= 19 3 =6.33

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**Section 5.3 – The Definite Integral**

Difference between the Value of a Definite Integral and Total Area Find the mean (average) value of the function 𝑓 𝑥 =𝑠𝑖𝑛𝑥 over the interval 0, 2𝜋 Total Area 0 2𝜋 𝑠𝑖𝑛𝑥 𝑑𝑥 = 0 𝜋 𝑠𝑖𝑛𝑥 𝑑𝑥 − 𝜋 2𝜋 𝑠𝑖𝑛𝑥 𝑑𝑥 = 𝜋 2𝜋 𝜋 Value of the Definite Integral −𝑐𝑜𝑠𝑥 − −𝑐𝑜𝑠𝑥 0 2𝜋 𝑠𝑖𝑛𝑥 𝑑𝑥 = 2𝜋 −𝑐𝑜𝑠𝑥 = −𝑐𝑜𝑠 𝜋 − −𝑐𝑜𝑠 0 − −𝑐𝑜𝑠 2𝜋 − −𝑐𝑜𝑠 𝜋 −𝑐𝑜𝑠 2𝜋 − −𝑐𝑜𝑠 0 = − −1 − −1 − − 1 − 1 = 4 − 1 − −1 = Copyright 2010 Pearson Education, Inc.

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