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Integrate pi·r^2·(4000r + 4000) from 0 to 1
\int_{0}^{1} \pi r^{2} \left(4000 r + 4000\right)\, dr
Step by step
- \int_{0}^{1} \pi r^{2} \left(4000 r + 4000\right)\, dr
First find an antiderivative F, then evaluate F(b) − F(a).
- \int \pi r^{2} \left(4000 r + 4000\right)\, dr = \pi \int r^{2} \left(4000 r + 4000\right)\, dr
Pull the constant \pi out of the integral.
- r^{2} \left(4000 r + 4000\right) = 4000 r^{3} + 4000 r^{2}
Rewrite the integrand into a friendlier form.
- \int 4000 r^{3} + 4000 r^{2}\, dr = \int 4000 r^{3}\, dr + \int 4000 r^{2}\, dr
The integral of a sum is the sum of the integrals.
- \int 4000 r^{3}\, dr = 4000 \int r^{3}\, dr
Pull the constant 4000 out of the integral.
- \int r^{3}\, dr = \frac{r^{4}}{4}
Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) (n ≠ −1).
- \int 4000 r^{2}\, dr = 4000 \int r^{2}\, dr
Pull the constant 4000 out of the integral.
- \int r^{2}\, dr = \frac{r^{3}}{3}
Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) (n ≠ −1).
- F(1) - F(0) = \left(\frac{7000 \pi}{3}\right) - \left(0\right)
Fundamental theorem of calculus: plug in the limits.
- = \frac{7000 \pi}{3} \approx 7330.4
Simplify.
Reveal the answer
\frac{7000 \pi}{3}