Find the values of a and b if:
∫2,1 xlnx dx = lna -(b/a)
btw, "∫2,1" means upper limit is 2, lower limit is 1
Help is very appreciated!!! Thanks!
Let $\displaystyle u'(x)=x~\implies~u(x)=\frac12 x^2$
and $\displaystyle v(x)=\ln(x)~\implies~v'(x)=\frac1x$
Your integral is then:
$\displaystyle \int x\cdot \ln(x) dx = \int u'(x) \cdot v(x) dx = u(x) \cdot v(x) - \int U(x) \cdot v'(x) dx = \frac12 x^2 \cdot \ln(x) - \int \frac12 x dx$
Finally you have:
$\displaystyle \int_1^2 x\cdot \ln(x) dx = \left[\frac12 x^2 \cdot \ln(x) - \frac14 x^2 \right]_1^2 = 2\ln(2) - \frac34 = \ln(4) - \frac34$
Compare the last term with the given term to determine the values aof a and b.
$\displaystyle \int_{1}^{2} x\ ln x dx $
du = x , u = x^2/2
v = ln x , dv = 1/x
$\displaystyle \int_{1}^{2} x\ln x \cdot dx = \frac{x^2 \cdot \ln x }{2}\mid_{1}^{2} - \int_{1}^{2} \frac{x^2}{2x} dx $
$\displaystyle \int_{1}^{2} x\ln x \cdot dx = 2\ln 2 - \ln 1 - \left( \frac{x^2}{4} \mid^2_1\right) $
$\displaystyle \int_{1}^{2} x\ln x \cdot dx = 2\ln 2 - 0 - \left( 1-\frac{1}{4}\right) $
$\displaystyle \int_{1}^{2} x\ln x \cdot dx = 2\ln 2 - \frac{3}{4} $
$\displaystyle \int_{1}^{2} x\ln x \cdot dx = \ln 2^2 - \frac{3}{4} $
a=4 , b=-3