How do you find #int_1^e 2x^2lnx #?

Question

Charles Arocha · Accepted Answer

You can use integration-by-parts to get #2/9+ 4/9 e^(3) approx 9.1491275#

Usually, when your integrand involves a logarithm, it's a good idea to try integration by parts with #u=ln(x)# and #du = 1/x dx#. In this case, we also get #dv = 2x^2 dx# so that #v=2/3 x^3#.

The integration-by-parts formula is #int u\ dv=uv- int v\ du#, so

#int 2x^2 ln(x)\ dx=2/3 x^{3}ln(x)-2/3 int x^2 dx#

#=2/3 x^{3}ln(x) - 2/9 x^{3}+C#.

Therefore, the definite integral is

#int_{1}^{e}2x^{2}ln(x)\ dx=(2/3 e^{3} * 1 - 2/9 e^{3}) - (0 - 2/9)#

#2/9+ 4/9 e^(3) approx 9.1491275#

Christopher Allers · Answer

#int_1^e 2x^2 ln(x) dx =2/9( 2e^3 + 1 )# #int_1^e 2x^2 ln(x) dx = 2/3 int_1^e frac{d}{dx}(x^3) ln(x) dx# #=2/3[ [x^3 ln(x)]_1^e - int_1^e x^3 frac{d}{dx}(ln(x)) dx ]# #=2/3[ (e^3 - 0) - int_1^e x^3 (1/x) dx ]# #=2/3[ e^3 - int_1^e x^2 dx ]# #=2/3[ e^3 - [x^3/3]_1^e ]# #=2/9[ 3e^3 - [x^3]_1^e ]# #=2/9[ 3e^3 - (e^3-1) ]# #=2/9( 2e^3 + 1 )#

Scarlett Allison · Answer

undefined To find $ \int_{1}^{e} 2x^2 \ln(x) $:

1. Use integration by parts.
2. Let $ u = \ln(x) $ and $ dv = 2x^2 \, dx $.
3. Calculate the differentials: $ du = \frac{1}{x} \, dx $ and $ v = \frac{2}{3}x^3 $.
4. Apply the integration by parts formula:
$$ \int u \, dv = uv - \int v \, du $$
5. Substitute the values:
$$ \int_{1}^{e} 2x^2 \ln(x) \, dx = \left[ \frac{2}{3}x^3 \ln(x) \right]_{1}^{e} - \int_{1}^{e} \frac{2}{3}x^3 \cdot \frac{1}{x} \, dx $$
$$ = \left[ \frac{2}{3}e^3 \ln(e) - \frac{2}{3}(1)^3 \ln(1) \right] - \int_{1}^{e} \frac{2}{3}x^2 \, dx $$
$$ = \frac{2}{3}e^3 - \frac{2}{3} + \left[ \frac{2}{9}x^3 \right]_{1}^{e} $$
$$ = \frac{2}{3}e^3 - \frac{2}{3} + \left( \frac{2}{9}e^3 - \frac{2}{9} \right) $$
$$ = \frac{2}{9}e^3 + \frac{2}{3} $$

Therefore, $ \int_{1}^{e} 2x^2 \ln(x) \, dx = \frac{2}{9}e^3 + \frac{2}{3} $.