Hard calc problem...any takers

Find the equation of the curve passing through (0, 1/2) if the slope is given by dy/dx = x4 + e2x.

if dy/dx = x^4 + e^(2x), then you just integrate this, and solve the new function using (0, 1/2) to find C.

⌠ x^4 + e^(2x) dx = (1/5)x^5 + (1/2)e^(2x) + C

Now use the point (0, 1/2) to find C.

(1/2) = (1/5)(0)^5 + (1/2)e^(2(0)) + C
(1/2) = 0 +1/2 + C
C = 0

Then the equation is

(1/5)x^5 + (1/2)e^(2x)

edit: technically JBrown is correct. Your integral is unclear. I solved the integral I believed you most likely intended to type. I might have been incorrect. Truly you need to ensure this is actually what you meant before you take my answer at face value.

edit: the answer below me should check his work before declaring mine is incorrect. (1/2)e^(2x) evaluated for x = 0, gives (1/2)e^0 = (1/2), not 1.

Use a carat (shift 6 on your keyboard) to indicate exponents. I assume your slope is

dy/dx = x^4 + e^(2x)

To find the original curve, y(x), we integrate both sides with respect to x.

∫dy/dx * dx = ∫(x^4 + e^(2x)) dx

The dx cancels on the left hand side, leaving ∫ dy = y plus a constant c1. On the right side, we can split up the integral.

y + c1 = ∫x^4 dx + ∫e^(2x) dx

The first term is easy to integrate using the power rule, ∫x^4 dx = x^5 / 5 + c2. We'll eventually incorporate the new constant of integration into the first one on the left side, but we may as well keep it around for now.

y + c1 = x^5 / 5 + c2 + ∫e^(2x) dx

For the second integral, let's define u = 2x. Then du/dx = 2, and dx = du / 2. Let's plug that in.

y + c1 = x^5 / 5 + c2 + ∫e^u du / 2

Now, we know that ∫e^u du = e^u + c3. We plug back in for u, and we get

y + c1 = x^5 / 5 + c2 + (e^(2x) + c3) / 2

Now, let's collect all the constants into a single value C.

y = x^5 / 5 + e^(2x) / 2 + (-c1 + c2 + c3/2)
y = x^5 / 5 + e^(2x) / 2 + C

Now, we have a "family of curves" which have a slope dy/dx = x^4 + e^(2x). Which one passes through (0, 1/2)? We have to evaluate the constant. Just plug in those x and y values.

1/2 = 0^5 + e^(2*0) / 2 + C
1/2 = 0 + e^0 / 2 + C
1/2 = 1/2 + C
C = 0

So, the curve that has the specified slope and passes through the specified point is

y = x^5 + e^(2x) + 0
y = x^5 + e^(2x)

I hope that helps!

U have to do integral to the derivative
x^4+ e^2x --------> (x^5)/5 + (e^2x)/2
so your function is
y=(x^5)/5 + (e^2x)/2
x=0 it gives you y=0.5 -----> makes sense

i would like clarification, is your equation x^4 + e^(2x) or x^4 + (e^2)x or is it something else?

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