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受欢迎的微积分 >

integral from 0 to-a of sqrt(x^2+a^2)

解答

\int_{0}^{- a} x^{2} + a^{2} d x

解答

- \frac{( 2 - ln ( - 1 + 2 ) ) a ^{2}}{2}

求解步骤

\int_{0}^{- a} x^{2} + a^{2} d x

使用三角换元法

= \int_{0}^{- \frac{π}{4}} a^{2} sec^{3} (u) d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= a^{2} \cdot \int_{0}^{- \frac{π}{4}} sec^{3} (u) d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= a^{2} (- \int_{- \frac{π}{4}}^{0} sec^{3} (u) d u)

使用降次积分法

= a^{2} (- ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{0} + \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{0} sec (u) d u))

使用积分常见定则:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= a^{2} (- ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{0} + \frac{1}{2} [ln ∣ tan (u) + sec (u) ∣]_{- \frac{π}{4}}^{0}))

化简

a^{2} (- ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{0} + \frac{1}{2} [ln ∣ tan (u) + sec (u) ∣]_{- \frac{π}{4}}^{0})) : - a^{2} ([\frac{1}{2} sec (u) tan (u)]_{- \frac{π}{4}}^{0} + \frac{1}{2} [ln ∣ tan (u) + sec (u) ∣]_{- \frac{π}{4}}^{0})

= - a^{2} ([\frac{1}{2} sec (u) tan (u)]_{- \frac{π}{4}}^{0} + \frac{1}{2} [ln ∣ tan (u) + sec (u) ∣]_{- \frac{π}{4}}^{0})

带入上下限计算后得:

\frac{1}{2}

带入上下限计算后得:

- ln (2 - 1)

= - a^{2} (\frac{1}{2} + \frac{1}{2} (- ln (2 - 1)))

化简

= - \frac{( 2 - ln ( - 1 + 2 ) ) a ^{2}}{2}

流行的例子

integral from-3 to 0 of (4x^{1/3})

\int_{- 3}^{0} (4 x^{\frac{1}{3}}) d x

integral from 1 to 2e of 4ln(x)

\int_{1}^{2 e} 4 ln (x) d x

integral from-1 to 1 of-x^2+2-x^2

\int_{- 1}^{1} - x^{2} + 2 - x^{2} d x

integral from 0 to 1 of (x^3)/(2(1+x^4))

\int_{0}^{1} \frac{x ^{3}}{2 ( 1 + x ^{4} )} d x

integral from-4 to 4 of x^3-16x

\int_{- 4}^{4} x^{3} - 16 x d x