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Integrate 2·pi·x·(4 - x^2) from 0 to 2
\int_{0}^{2} 2 \pi x \left(4 - x^{2}\right)\, dx
Passo dopo passo
- \int_{0}^{2} 2 \pi x \left(4 - x^{2}\right)\, dx
First find an antiderivative F, then evaluate F(b) − F(a).
- \int 2 \pi x \left(4 - x^{2}\right)\, dx = 2 \pi \int x \left(4 - x^{2}\right)\, dx
Pull the constant 2 \pi out of the integral.
- u = x^{2},\quad du = 2 x\, dx
Substitute u = x^{2}.
- \int x \left(4 - x^{2}\right)\, dx = \int 2 - \frac{u}{2}\, d_u
Rewrite the integral in terms of u.
- \int 2 - \frac{u}{2}\, d_u = \int 2\, d_u + \int - \frac{u}{2}\, d_u
The integral of a sum is the sum of the integrals.
- \int 2\, d_u = 2 u
The integral of a constant c is c·x.
- \int - \frac{u}{2}\, d_u = - \frac{1}{2} \int u\, d_u
Pull the constant - \frac{1}{2} out of the integral.
- \int u\, d_u = \frac{u^{2}}{2}
Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) (n ≠ −1).
- = - \frac{x^{4}}{4} + 2 x^{2}
Substitute back u = x^{2}.
- F(2) - F(0) = \left(8 \pi\right) - \left(0\right)
Fundamental theorem of calculus: plug in the limits.
- = 8 \pi \approx 25.133
Simplify.
Rivela la risposta
8 \pi