Find area between $3$ functions

Question

Find the area between the curve $y=4-x^2$, the tangent line of the parabola at $x=1$ and the vertical asymptote of $f(x)=x\ln{(x-3)}$.

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The tangent line of the parabola at $x=1$ is $$y=-2x+5.$$ The V.A. of $f$ is $$x=3,$$ so I think the area to look for is

So I did $$\int_1^3{\left[(-2x+5)-\left(4-x^2\right)\right]}\;\text dx=\cdots=\boxed{\frac 83},$$ nevertheless WolframAlpha says other thing ($\frac{16}3$).

What am I doing wrong?

Thank you!

Answer 1

Wolfram sets the left border itself by solving the equation. For example:

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The left border: $5-2x=3 \ \text{or} \ 4-x^2=3 \Rightarrow x=1 \ \text{or} \ x=\pm 1 \Rightarrow x=-1.$

Another example:

$\hspace{3cm}$

The left border: $5-2x=4 \ \text{or} \ 4-x^2=4 \Rightarrow x=\frac12 \ \text{or} \ x=0 \Rightarrow x=0.$

Conclusion: Show the left and right borders to avoid misunderstanding.

Answer 2

Following your link I see that W-A shows:

$\int_{-1}^1 (1 - 2 x + x^2) dx + \int_1^3 (1 - 2 x + x^2) dx = 16/3≈5.33333$

But I don't see why it wants to calculate over the interval [-1,1].