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用定义求3/π到0的定积分cosxdx,为什么用定义求和用牛顿莱布尼茨公式求的答案符号相反呢?
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用定义求3/π 到0的定积分cosxdx,为什么用定义求和用牛顿莱布尼茨公式求的答案符号相反呢?
▼优质解答
答案和解析
By formula:
∫(π/3,0) cosx dx
= -sinx
= -√3/2
By definition:
The nth Area = Σ(k=1->n) cos[π/3 + (-kπ/3)/n] * (-π/3)/n
= [-π/(3n)]Σ(k=1->n) cos[π/3-(kπ)/(3n)]
= [-π/(3n)] * (1/4){√3 * cot[π/(6n)] + 1}
= -(π/12) * {√3 * cot[π/(6n)] + 1}/n
A = ∫(π/3,0) cosx dx
= lim(n->∞) -(π/12) * {√3 * cot[π/(6n)] + 1}/n
= (-π/12)lim(n->∞) {1 + √3/tan[π/(6n)]}/n
= (-π/12){[lim(n->∞) 1/n] + [√3lim(n->∞) 1/(n*tan(π/(6n))]}
= (-π√3/12)lim(n->∞) 1/{n*tan[π/(6n)]}
= (-π√3/12)lim(n->∞) 6/{π[tan²(π/(6n))+1]},L’Hospital's Rule
= (-√3/2)lim(n->∞) 1/{1+tan²[π/(6n)]}
= (-√3/2) * 1/(1+0)
= -√3/2
∫(π/3,0) cosx dx
= -sinx
= -√3/2
By definition:
The nth Area = Σ(k=1->n) cos[π/3 + (-kπ/3)/n] * (-π/3)/n
= [-π/(3n)]Σ(k=1->n) cos[π/3-(kπ)/(3n)]
= [-π/(3n)] * (1/4){√3 * cot[π/(6n)] + 1}
= -(π/12) * {√3 * cot[π/(6n)] + 1}/n
A = ∫(π/3,0) cosx dx
= lim(n->∞) -(π/12) * {√3 * cot[π/(6n)] + 1}/n
= (-π/12)lim(n->∞) {1 + √3/tan[π/(6n)]}/n
= (-π/12){[lim(n->∞) 1/n] + [√3lim(n->∞) 1/(n*tan(π/(6n))]}
= (-π√3/12)lim(n->∞) 1/{n*tan[π/(6n)]}
= (-π√3/12)lim(n->∞) 6/{π[tan²(π/(6n))+1]},L’Hospital's Rule
= (-√3/2)lim(n->∞) 1/{1+tan²[π/(6n)]}
= (-√3/2) * 1/(1+0)
= -√3/2
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