calculusintegration
Find the Area of the shaded region
$y= \ln x \rightarrow x = e^y $
I found that the curve cuts the $x$ axis at $1$ through substituting $0$ to $y$
Why I can't find the area under the graph through this and how should I find it ?
$$ \int_1^4 e^y \, dy $$
Hints:
$x_B$ can be found by solving $3x-x^2=0$. Now you can work out the area of the right triangle colored red in the figure.
Now the area of the shaded region can be represented as
$$A=\int_{x_A}^{x_B} 3x-x^2\,dx-\text{area of the triangle}$$
First evaluate the x value of $B$.
$$3x-x^2=-3x$$ $$6x-x^2=0$$ $$x(6-x)=0$$ So $B$ is at $x=6$ (and $A$ is at $x=0$)
Construct integral: $\int_0^63x-x^2\space-\space-3x\space dx$
$$\int_0^63x-x^2\space-\space-3x\space dx=\int_0^6 6x-x^2dx$$ $$=\left(3x^2-\frac{x^3}{3}\right)_0^6$$ $$=3\times6^2-\frac{6^3}{3}$$ $$=108-72$$ $$=36$$
Best Answer
Way 1
By part $$\int_1^4 \ln(x)\,dx=\int_1^4 \underbrace{1}_{u'(x)}\underbrace{\ln(x)}_{v(x)} \, dx= \left[ x \ln(x) \vphantom{\frac 11} \right]_1^4-\int_1^4dx=\cdots$$
Way 2
If you make the substitution $x=e^u$, then you get $$\int_1^4 \ln(x)\,dx=\int_{\ln(1)}^{\ln(4)}ue^u\,du.$$