integral from 1 to 5 of 7r^2ln(r) — step-by-step solution

⚠ 7r is read as the product 7*r: for a power, write 7^r

1) Let's compute the integral:

\int(7 r^{2} \cdot \ln (r))dr

For

7 r^{2} \cdot \ln (r)

, we use integration by parts:

\int(u1 \cdot v2')dr = u1 \cdot v2 - \int(u1' \cdot v2)dr

, with u1 =

\ln (r)

and v2' =

r^{2}

2) Let's perform the differentiation u1:

\frac{d}{d r} (\ln (r))

\frac{1}{r}

3) Let's compute the integral v2':

\int(r^{2})dr

\frac{r^{3}}{3}

4) Let's calculate

u1' \cdot v2

\frac{1}{r} \cdot \frac{r^{3}}{3}

ℹ

\frac{1}{r} \cdot \frac{r^{3}}{3}

\frac{r^{3}}{r \cdot 3}

‖

\frac{r^{3}}{r \cdot 3}

\frac{r^{2}}{3}

\frac{r^{2}}{3}

5) Let's compute the integral

u1 \cdot v2 - \int(u1' \cdot v2)dr

7 \cdot (\ln (r) \cdot \frac{r^{3}}{3} - \int(\frac{r^{2}}{3})dr)

ℹ

\ln (r) \cdot \frac{r^{3}}{3}

\frac{\ln (r) \cdot r^{3}}{3}

‖

- \int(\frac{r^{2}}{3})dr

- \frac{1}{3} \cdot \frac{r^{3}}{3}

7 \cdot (\frac{\ln (r) \cdot r^{3}}{3} - \frac{1}{3} \cdot \frac{r^{3}}{3})

7 \cdot (\frac{\ln (r) \cdot r^{3}}{3} - \frac{r^{3}}{9})

7 \cdot (\frac{3 \cdot \ln (r) \cdot r^{3} - r^{3}}{9})

▷

\frac{7 r^{3} \cdot (3 \cdot \ln (r) - 1)}{9}

6) Let's evaluate for the interval

[1, 5]

\frac{7 \cdot 5^{3} \cdot (3 \cdot \ln (5) - 1)}{9} - \frac{7 \cdot 1^{3} \cdot (3 \cdot \ln (1) - 1)}{9}

\frac{7 \cdot 125 \cdot (3 \cdot \ln (5) - 1)}{9} - \frac{7 \cdot (3 \cdot \ln (1) - 1)}{9}

ℹ

7 \cdot 125

875

‖

\ln (1)

0

\frac{875 \cdot (3 \cdot \ln (5) - 1)}{9} - \frac{7 \cdot (3 \cdot 0 - 1)}{9}

\frac{875 \cdot (3 \cdot \ln (5) - 1)}{9} - \frac{7 \cdot (0 - 1)}{9}

\frac{875 \cdot (3 \cdot \ln (5) - 1)}{9} - \frac{-7}{9}

\frac{875 \cdot (3 \cdot \ln (5) - 1)}{9} + \frac{7}{9}

\frac{875 \cdot (3 \cdot \ln (5) - 1) + 7}{9}

\frac{(2625 \cdot \ln (5) - 875) + 7}{9}

▶Integral:

\frac{2625 \cdot \ln (5) - 868}{9}

7) The numeric value of:

\frac{2625 \cdot \ln (5) - 868}{9}

is:

\frac{2625 \cdot \approx 1.6094 - 868}{9}

\frac{\approx 4224.7745 - 868}{9}

\frac{\approx 3356.7745}{9}

▷

\approx 372.9749