Si(x)=∫0xsin(t)tdtSi(𝑥)=∫0𝑥sin(𝑡)𝑡𝑑𝑡 near the point A=[3,Si(3)]≈[3,1.849]𝐴=[3,Si(3)]≈[3,1.849]. The differential approximation to SiSi near 33 is given by the formulaSi(x)≈1.849+Si′(3)(x−3)Si(𝑥)≈1.849+Si′(3)(𝑥−3). Since Si′(3)=Si′(3)= , we have the linear approximation to Si(x)Si(𝑥) near x=3𝑥=3:
Question
Si(x)=∫0xsin(t)tdtSi(𝑥)=∫0𝑥sin(𝑡)𝑡𝑑𝑡 near the point A=[3,Si(3)]≈[3,1.849]𝐴=[3,Si(3)]≈[3,1.849]. The differential approximation to SiSi near 33 is given by the formulaSi(x)≈1.849+Si′(3)(x−3)Si(𝑥)≈1.849+Si′(3)(𝑥−3). Since Si′(3)=Si′(3)= , we have the linear approximation to Si(x)Si(𝑥) near x=3𝑥=3:
Solution
Para encontrar la aproximación lineal de cerca de , primero necesitamos calcular la derivada de en .
Dado que , podemos usar el teorema fundamental del cálculo para encontrar la derivada:
Ahora evaluamos esta derivada en :
Aproximadamente, , entonces:
Con esta información, podemos escribir la aproximación lineal de cerca de :
Por lo tanto, la aproximación lineal de cerca de es:
Similar Questions
Evaluating integrals can be difficult. So let's instead use the total differential approximation to approximate the Sine IntegralSi(x)=∫0xsin(t)tdtSi(𝑥)=∫0𝑥sin(𝑡)𝑡𝑑𝑡 near the point A=[3,Si(3)]≈[3,1.849]𝐴=[3,Si(3)]≈[3,1.849] . The differential approximation to SiSi near 33 is given by the formulaSi(x)≈1.849+Si′(3)(x−3)Si(𝑥)≈1.849+Si′(3)(𝑥−3) . Since Si′(3)=Si′(3)= , we have the linear approximation to Si(x)Si(𝑥) near x=3𝑥=3 : Si(x)≈Si(𝑥)≈ . Use the linear approximation to evaluate (to 3 decimal places) Si(2.9)≈Si(2.9)≈ IncorrectYour Answer: 1.84 .
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