# Integration It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution:

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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 ’

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!

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!

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!

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

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!

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

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!

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

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

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

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

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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