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# Exercicios de integrais

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Resolução de integrais trigonométricas

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### Exercicios de integrais

1. 1. INTEGRATION BY SUBSTITUTION Section 5.5 November 25, 2013
2. 2. LAST TIME By the FTC (Part I): If f is continuous on [a, b] and F(x) = x a f(t) dt =⇒ F (x) = f(x) F(x) = a x f(t) dt =⇒ F (x) = − f(x) F(x) = g(x) a f(t) dt =⇒ F (x) = f(g(x)) · g (x)
3. 3. SUBSTITUTION Substitution is a notational trick to make it easier to determine indeﬁnite integrals. It uses something called a differential. Deﬁnition Let u = f(x) for some function f. The differential of u, denoted du, is given by du = f (x) dx. For example, u = 7x2 + 1 =⇒ du = 14x dx u = 24x =⇒ du = 4 ln(2)24x dx For us, this is purely notational.
4. 4. SUBSTITUTION Determine (3x2 − 5)4 2x dx. Let u = 3x2 − 5. Then du = 6x dx =⇒ dx = du 6x Rewriting the above indeﬁnite integral using u and du we get (3x2 − 5)4 2x dx = u4 2x du 6x = 1 3 u4 du This indeﬁnite integral is easier to determine: 1 3 u4 du = 1 3 u4 du = 1 3 1 5 u5 + C = 1 15 u5 + C Since u = 3x2 − 5, substituting back we get: 1 15 (3x2 − 5)5 + C
5. 5. SUBSTITUTION What did we do? (1) We chose a u and determined du. (2) We rewrote our integral using u and du. (3) Found the indeﬁnite integral. (4) Substitute back for u. Question: How do we choose our u? No foolproof way, but in general, choose: the term under a root or raised to a power, the exponent of an exponential function, the logarithm, the term in the denominator, the term inside a trig function.
6. 6. EXAMPLES Determine x 1 − x2 dx. u = 1 − x2 du = − 2x dx (dx = −du/2x) Rewriting the above indeﬁnite integral using u and du we get x 1 − x2 dx = x √ u −1 2x du = − 1 2 √ u du This indeﬁnite integral is easier to determine: − 1 2 u1/2 du = − 1 2 1 1 2 + 1 u 1 2 +1 + C = − 1 3 u3/2 + C Since u = 1 − x2, we get: − 1 3 (1 − x2 )3/2 + C
7. 7. EXAMPLES Determine 3t2 e2t3 dt. u = 2t3 du = 6t2 dt (dt = du/6t2) Rewriting the above indeﬁnite integral using u and du we get 3t2 e2t3 dt = 3t2 eu 1 6t2 du = 1 2 eu du This indeﬁnite integral is easier to determine: 1 2 eu du = 1 2 eu + C Since u = 2t3, we get: 1 2 e2t3 + C
8. 8. EXAMPLES Determine 1 x ln(x) dx. u = ln(x) du = 1 x dx (dx = xdu) Rewriting the above indeﬁnite integral using u and du we get 1 x ln(x) dx = 1 xu xdu = 1 u du This indeﬁnite integral is easier to determine: 1 u du = ln |u| + C Since u = ln(x), we get: ln | ln(x)| + C
9. 9. EXAMPLES Determine y2 2y3 + 1 dy. u = 2y3 + 1 du = 6y2 dy (dy = du/6y2) Rewriting the above indeﬁnite integral using u and du we get y2 2y3 + 1 dy = y2 u · 1 6y2 du = 1 6 1 u du This indeﬁnite integral is easier to determine: 1 6 1 u du = 1 6 ln |u| + C Since u = 2y3 + 1, we get: 1 6 ln |2y3 + 1| + C
10. 10. EXAMPLES Determine t sin(2t2 ) dt. u = 2t2 du = 4t dt (dt = du/4t) Rewriting the above indeﬁnite integral using u and du we get t sin(2t2 ) dt = t sin(u) 1 4t du = 1 4 sin(u) du This indeﬁnite integral is easier to determine: 1 4 sin(u) du = − 1 4 cos(u) + C Since u = 2t2, we get: − 1 4 cos(2t2 ) + C
11. 11. PRACTICE PROBLEMS Use substitution to determine the following indeﬁnite integrals: (1) z(4z2 − 5)4 dz (2) (1 − t)e2t−t2 dt (3) r r2 + 2 dr (4) θ2 sec2 (5θ3 + 10) dθ
12. 12. PRACTICE PROBLEMS (1) z(4z2 − 5)4 dz. u = 4z2 − 5 du = 8z dz (dz = du/8z) Rewriting the above indeﬁnite integral using u and du we get z(4z2 − 5)4 dz = zu4 1 8z du = 1 8 u4 du This indeﬁnite integral is easier to determine: 1 8 u4 du = 1 40 u5 + C Since u = 4z2 − 5, we get: 1 40 (4z2 − 5)5 + C
13. 13. PRACTICE PROBLEMS (2) (1 − t)e2t−t2 dt. u = 2t − t2 du = (2 − 2t) dt (dt = du/(2 − 2t)) Rewriting the above indeﬁnite integral using u and du we get (1 − t)e2t−t2 dt = (1 − t)eu 1 2 − 2t du = 1 2 eu du This indeﬁnite integral is easier to determine: 1 2 eu du = 1 2 eu + C Since u = 2t − t2, we get: 1 2 e2t−t2 + C
14. 14. PRACTICE PROBLEMS (3) r r2 + 2 dr. u = r2 + 2 du = 2r dr (dr = du/2r) Rewriting the above indeﬁnite integral using u and du we get r r2 + 2 dr = r u 1 2r du = 1 2 · 1 u du This indeﬁnite integral is easier to determine: 1 2 1 u du = 1 2 ln |u| + C Since u = r2 + 2, we get: 1 2 ln |r2 + 2| + C
15. 15. PRACTICE PROBLEMS (4) θ2 sec2 (5θ3 + 10) dθ. u = 5θ3 + 10 du = 15θ2 dθ (dθ = du/15θ2) Rewriting the above indeﬁnite integral using u and du we get θ2 sec2 (5θ3 + 10) dθ = θ2 sec2 (u) 1 15θ2 du = 1 15 sec2 (u) du This indeﬁnite integral is easier to determine: 1 15 sec2 (u) du = 1 15 tan(u) + C Since u = 5θ3 + 10, we get: 1 15 tan(5θ3 + 10) + C
16. 16. SUBSTITUTION It may be the case that after substituting for u and du the original variable remains in the integrand. Determine x(5x − 1)2/3 dx u = 5x − 1 du = 5 dx (dx = du/5) x(5x − 1)2/3 dx = 1 5 xu2/3 du What do we do about the x? Try to rewrite it in terms of u, u = 5x − 1 ⇒ x = (u + 1)/5. 1 5 xu2/3 du = 1 25 (u + 1)u2/3 du = 1 25 u5/3 + u2/3 du 3 200 u8/3 + 3 125 u5/3 + C = 3 200 (5x − 1)8/3 + 3 125 (5x − 1)5/3 + C
17. 17. EXAMPLE Determine p3 (p2 + 1)5 dp. u = p2 + 1 du = 2p dp (dp = du/2p) Rewriting the above indeﬁnite integral using u and du we get p3 2p u5 du = 1 2 p2 u5 du Since u = p2 + 1, we know p2 = u − 1 1 2 p2 u5 du = 1 2 (u − 1)u5 du = 1 2 u6 − u5 du = 1 14 u7 − 1 12 u6 + C Since u = p2 + 1, we get: 1 14 (p2 + 1)7 − 1 12 (p2 + 1)6 + C