The Chemistry Maths Book, Second Edition

150 Chapter 5Integration whereR 1 = 1 r 1 1 + 1 r 2 is the distance between the masses. The moment of inertia is then where the ...

5.6 Static properties of matter 151 The position of the centre of mass is given by (see equation (5.40)) (5.46) and the moment o ...

152 Chapter 5Integration 5.7 Dynamics Velocity and distance As in Section 5.1, we consider a body moving along a curve from poin ...

5.7 Dynamics 153 EXAMPLE 5.16A body falling under the influence of gravity A body of mass mfalling freely under the influence of ...

154 Chapter 5Integration This work is positive for like charges. The same formula applies to the case of unlike charges, but the ...

5.7 Dynamics 155 dissipative forces such as those due to friction. Then, by the fundamental theorem of the calculus (equation (5 ...

156 Chapter 5Integration We note that, whilst the kinetic energy has a well defined absolute value, this is not true of the pote ...

5.8 Pressure–volume work 157 (where the potential energy is zero). At an intermediate height,E 1 = 1 mgh 1 = 1 T(x) 1 + 1 V(x) a ...

158 Chapter 5Integration according as the external pressurep ext is greater than or less than p. Ifp ext 1 > 1 pthen the flui ...

5.8 Pressure–volume work 159 The total work done on the fluid in the cyclea 1 → 1 b 1 → 1 ais therefore W aba 1 = 1 W ab 1 + 1 W ...

160 Chapter 5Integration 5.9 Exercises Section 5.2 Evaluate the indefinite integrals: Eval ...

5.9 Exercises 161 Find the average values in the given intervals: 2 x 2 1 + 13 x 1 + 1 4; − 11 ≤ 1 x 1 ≤ 1 + 1 27.cos 13 θ;0 1 ...

162 Chapter 5Integration (iv)the moment of inertia with respect to an arbitrary point x 0 on the line, (v)the moment of inertia ...

6 Methods of integration 6.1 Concepts To calculate the value of a definite integral it is normally necessary to evaluate the cor ...

164 Chapter 6Methods of integration (6.3) (6.4) (6.5) (6.6) More generally, the relations can be used to express a functionsin m ...

6.3 The method of substitution 165 Integral 3: By equation (6.3), Therefore where C′is a new arbitrary constant. Most tabulation ...

166 Chapter 6Methods of integration f(x) 1 = 1 (2x 1 − 1 1) 3 1 = 18 x 3 1 − 112 x 2 1 + 16 x 1 − 11 = 12 x 4 1 − 14 x 3 1 + 13 ...

6.3 The method of substitution 167 and, substituting in (6.7), (6.9) The essential skill in applying the method of substitution ...

168 Chapter 6Methods of integration EXAMPLE 6.4Integrate. Letu 1 = 12 x 2 1 + 13 x 1 + 11. Thendu 1 = 1 (4x 1 + 1 3)dx,and 0 Exe ...

6.3 The method of substitution 169 (ii). Because , we letu 1 = 1 ln 1 xand Then 0 Exercises 17–20 EXAMPLES 6.6Integrals of type ...