- #1

songoku

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- Homework Statement
- Given the equation of the graph in parametric form:

##x=t^2+1##

##y=3+3t##

Normal at point P (5, 9) cuts x-axis at Q. Find area R

- Relevant Equations
- Integration

I want to ask about the solution. The solution divides region R into two parts: curved part and triangle. The triangle is obtained by drawing line ##x=5##. Let say line ##x=5## cuts x-axis at point A so the triangle is PAQ

For the curved part:

$$\int_{-1}^{2} (3+3t) ~2t~ dt$$

My question:

Why the limit of integration for curved part is from ##t=-1## to ##t=2##? The value of ##x## at the leftmost part of the graph is ##x=1## so why the limit is not from ##t=0## to ##t=2##?

In my imagination, the curved area covered by the solution is only from ##x=2## to ##x=5## so there are region R from ##x=1## to ##x=2## that is not covered.

Thanks