MATH 12002 - CALCULUS I 5.3: Integrals and the Natural Exponential - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I § 5.3: Integrals and the Natural Exponential Function Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 7

Integral of e x Recall the differentiation formula d dx e x = e x . Reversing this formula gives the integration formula � e x dx = e x + C . D.L. White (Kent State University) 2 / 7

Examples Evaluate the integrals: � x 3 e x 4 +5 dx . 1 � � x 3 e x 4 +5 dx x 3 e x 4 +5 dx = Let u = x 4 + 5 so du = 4 x 3 dx 4 du = x 3 dx and 1 1 � e u du = 4 1 4 e u + C = 1 4 e x 4 +5 + C . = Note: This integral can also be evaluated by letting u = e x 4 +5 , so that du = 4 x 3 e x 4 +5 dx and 1 4 du = x 3 e x 4 +5 dx . The integral becomes � 1 4 e x 4 +5 + C . 4 du = 1 4 u + C = 1 D.L. White (Kent State University) 3 / 7

Examples � e − 5 x dx . 2 We can use a formal substitution, letting u = − 5 x , so that du = − 5 dx and − 1 5 du = dx , and so e − 5 x dx = − 1 e u du = − 1 5 e u + C = − 1 � � 5 e − 5 x + C . 5 However, when u is just a constant multiple of x , it is usually more efficient to use the “guess and check” method: e − 5 x is approximately e − 5 x . � We would guess that dx e − 5 x = − 5 e − 5 x d Now check the derivative: and so we don’t get the function we’re integrating. So we multiply our “guess” by − 1 5 to compensate, and check again: dx ( − 1 5 e − 5 x ) = − 1 d 5 ( − 5 e − 5 x ) = e − 5 x . 5 e − 5 x + C . e − 5 x dx = − 1 dx ( − 1 d 5 e − 5 x ) = e − 5 x , we have that � Since D.L. White (Kent State University) 4 / 7

Examples � ln π e x cos( e x ) dx . 3 ln π 2 � ln π � ln π e x cos( e x ) dx e x cos( e x ) dx = ln π ln π 2 2 u = e x du = e x dx x = ln π ⇒ u = e ln π = π x = ln π 2 ⇒ u = e ln π 2 = π 2 π � π � � = cos u du = sin u � � π � 2 π/ 2 sin π − sin π = 2 = 0 − 1 = − 1 . D.L. White (Kent State University) 5 / 7

Examples e x � (1 + e x ) 2 dx . 4 e x � � e x (1 + e x ) − 2 dx (1 + e x ) 2 dx = Let u = 1 + e x , so du = e x dx � 1 u − 2 du = − u − 1 + C = − = 1 + e x + C . e x � 1 + e x dx . 5 e x e x � � 1 + e x dx = 1 + e x dx Let u = 1 + e x , so du = e x dx � 1 u du = ln | u | + C = ln(1 + e x ) + C . = D.L. White (Kent State University) 6 / 7

Examples � 1 + e x dx . 6 e x � 1 + e x e x + e x � 1 = dx e x dx e x � e − x + 1 dx = − e − x + x + C . = � e dx . 7 Since e is a constant, � e dx = ex + C , � � just like 3 dx = 3 x + C or − 735 dx = − 735 x + C . Be sure to distinguish between the constant e ≈ 2 . 718281828459045 and the exponential function e x . D.L. White (Kent State University) 7 / 7