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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

  1. \int_{0}^{1} \pi r^{2} \left(4000 r + 4000\right)\, dr

    First find an antiderivative F, then evaluate F(b) − F(a).

  2. \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.

  3. r^{2} \left(4000 r + 4000\right) = 4000 r^{3} + 4000 r^{2}

    Rewrite the integrand into a friendlier form.

  4. \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.

  5. \int 4000 r^{3}\, dr = 4000 \int r^{3}\, dr

    Pull the constant 4000 out of the integral.

  6. \int r^{3}\, dr = \frac{r^{4}}{4}

    Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) (n ≠ −1).

  7. \int 4000 r^{2}\, dr = 4000 \int r^{2}\, dr

    Pull the constant 4000 out of the integral.

  8. \int r^{2}\, dr = \frac{r^{3}}{3}

    Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) (n ≠ −1).

  9. F(1) - F(0) = \left(\frac{7000 \pi}{3}\right) - \left(0\right)

    Fundamental theorem of calculus: plug in the limits.

  10. = \frac{7000 \pi}{3} \approx 7330.4

    Simplify.

Reveal the answer
\frac{7000 \pi}{3}