# Lecture 7 Substitution and Area Between Curves - W15 MAT...

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W15 MAT 21B LECTURE 7: SUBSTITUTION AND AREA BETWEEN CURVES GEORGE MOSSESSIAN Definite integration. So we’ve now talked about using substitution to compute indefinite integrals. Going from indefinite integrals to definite ones is pretty straightforward, but when you’re using substitution, there’s something you have to be careful of. Let’s look at an example: Example 1. Condsider Z 1 - 1 3 x 2 p x 3 + 1 dx. Now, when we substitute u = x 3 + 1, the integrand (the stuff inside the integral) becomes udu . But remember, the integral is a computation of area under the graph of the integrand, between two points x = a and x = b . But our new integral is no longer an x integral, it’s a u integral. So when x = - 1, u = ( - 1) 3 + 1 = 0, and when x = 1, u = (1) 3 + 1 = 2. Therefore the definite integral has become Z 2 0 udu = 2 3 u 3 / 2 2 0 = 4 2 3 . Of course, you don’t have to do this. You can also convert the antiderivative back to being a function of x , like so: Z x =1 x = - 1 udu = 2 3 u 3 / 2 x =1 x = - 1 = 2 3 ( x 3 + 1) 3 / 2 1 - 1 = 2 3 h (1 3 + 1) 3 / 2 - (( - 1) 3 + 1) 3 / 2 i = 4 2 3 . Example 2. Z π/ 2 π/ 4 cot θ csc 2 θdθ, let u = cot θ , so - du = csc 2 θdθ , and u ( θ = π/ 4) = 1, u ( θ = π/ 2) = 0. Then the integral becomes - Z 0 1 udu = Z 1 0 udu = u 2 2 u =1 u =0 = 1 2 Exercise 1. Do this the other way, by converting the integral back to θ and plugging in the original values. Even and odd functions. Recall that f ( x ) is called odd if f ( - x ) = - f ( x ), and f ( x ) is even if f ( - x ) = f ( x ). Visually, a function is odd if the graph has the property that, when you spin it 180 degrees around