Equation of a curve passing through (1,2)

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Homework Statement



A curve has slope 2x + 3 at each point (x, y) on the curve. Which of the following is an equation for this curve if it pases through the point (1,2)?

Answer: y= x2 + 3x - 2

Homework Equations



y= mx + b

The Attempt at a Solution



If the slope is 2x + 3, then that means it is the derivative. Since I a point (1,2), I can find m and b.

m = 2(1) + 3 = 5​

Thus,

2 = 5(1) + b
b = -3​

Therefore, my equation is

y= 5x - 3​

But this isn't correct. What am I doing wrong?
 
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You found the equation of the tangent to the curve at (1,2), not the equation of the curve.
 
vela said:
You found the equation of the tangent to the curve at (1,2), not the equation of the curve.

i agree.
To find the equation of the curve you will be doing some integration.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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