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Integration It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: To find: 6F To integrate this, you need to replace the x terms with equivalent u terms, and replace the dx with an equivalent du DifferentiateRearrange to find x Rearrange to get dx Replace each ‘x’ term with an equivalent ‘u’ term Rearrange – you should leave ‘du’ at the end Combine terms including the square root, changed to a power ‘ 1 / 2 ’

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Integration It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: To find: 6F Differentiate terms separately Flip the dividing fractions Calculate the fraction parts Finally, replace with u with its equivalent from the start!

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Integration It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: To find: 6F DifferentiateRearrange to find Sinx Rearrange to get dx Replace each ‘x’ term with an equivalent ‘u’ term Cancel the Cosx terms Multiply out Integrate Replace u with x terms again!

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Integration It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use integration by substitution to find: Sometimes you will have to decide on a substitution yourself. In this case, the bracket would be hardest to integrate so it makes sense to use the substitution: 6F Differentiate Rearrange to find x Rearrange to get dx You also need to recalculate limits in terms of u Replace x limits with u limits and the x terms with u terms Multiply out bracket Integrate Sub in limits and calculate An alternative method is to replace the ‘u’ terms with x terms at the end and then just use the original ‘x’ limits – either way is fine!

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Integration You can use integration by parts to integrate some expressions You may need the following Integrals, which you are given in the formula booklet… 6G

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Integration You can use Integration to find areas and volumes 6I x y a b You already know how to find the area under a curve by Integration Imagine we rotated the area shaded around the x-axis What would be the shape of the solid formed? y x This would be the solid formed In this section you will learn how to find the volume of any solid created in this way. It also involves Integration!

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Integration You can use Integration to find areas and volumes 6J x y ab In the trapezium rule we thought of the area under a curve being split into trapezia. To simplify this explanation, we will use rectangles now instead The height of each rectangle is y at its x-coordinate The width of each is dx, the change in x values So the area beneath the curve is the sum of ydx (base x height) The EXACT value is calculated by integrating y with respect to x (y dx) y dx y x For the volume of revolution, each rectangle in the area would become a ‘disc’, a cylinder The radius of each cylinder would be equal to y The height of each cylinder is dx, the change in x So the volume of each cylinder would be given by πy 2 dx The EXACT value is calculated by integrating y 2 with respect to x, then multiplying by π. ab y dx

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Integration You can use Integration to find areas and volumes The volume of revolution of a solid rotated 2π radians around the x-axis between x = a and x = b is given by: 1) The region R is bounded by the curve y = sin2x, the x-axis and the vertical lines x = 0 and x = π / 2. Find the volume of the solid formed when the region is rotated 2π radians about the x-axis. 6J Sub in a, b and y Square the bracket Using the identity Cos2A = 1 – 2sin 2 A, replace sin 2 2x with something equivalent The 1 / 2 can be put outside the integral Integrate and use a square bracket with the limits Sub in the two limits And finally we have the volume!

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Integration You can use Integration to find areas and volumes The volume of revolution can also be calculated when x and y are given parametrically. In this case you must also include dx / dt in the integral. You will also need to change limits so they are in terms of t rather than x! 6J

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Integration You can use Integration to find areas and volumes The volume of revolution can also be calculated when x and y are given parametrically. In this case you must also include dx / dt in the integral. The curve C has parametric equations: Where t ≥ 0. The region R is bounded by C, the x- axis and the lines x = 0 and x = 2. Find the volume of the solid formed when R is rotated 2π radians about the x-axis. 6J t = -1 t = 0 We know A = -1 from before Replace y, and calculate dx / dt Square the bracket and combine them We need to use partial fractions here Combine with a common denominator Sub in values to find A and B

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Integration You can use Integration to find areas and volumes The volume of revolution can also be calculated when x and y are given parametrically. In this case you must also include dx / dt in the integral. The curve C has parametric equations: Where t ≥ 0. The region R is bounded by C, the x- axis and the lines x = 0 and x = 2. Find the volume of the solid formed when R is rotated 2π radians about the x-axis. 6J Replace y, and calculate dx / dt Square the bracket and combine them We need to use partial fractions here We also need to calculate the limits for t rather than x Sub in the x limits and solve for t x = 0x = 2

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Integration You can use Integration to find areas and volumes The volume of revolution can also be calculated when x and y are given parametrically. In this case you must also include dx / dt in the integral. The curve C has parametric equations: Where t ≥ 0. The region R is bounded by C, the x- axis and the lines x = 0 and x = 2. Find the volume of the solid formed when R is rotated 2π radians about the x-axis. 6J Replace y, and calculate dx / dt Square the bracket and combine them We need to use partial fractions here Use t-limits

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Integration You can use Integration to find areas and volumes The volume of revolution can also be calculated when x and y are given parametrically. In this case you must also include dx / dt in the integral. The curve C has parametric equations: Where t ≥ 0. The region R is bounded by C, the x- axis and the lines x = 0 and x = 2. Find the volume of the solid formed when R is rotated 2π radians about the x-axis. 6J Integrate and write as a square bracket Sub in limits separately Simplify/Calculate

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Chapter 6 – Applications of Integration

Chapter 6 – Applications of Integration

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