Introduction to Integrals

How to find a function from its derivative and a point?

Quick answer

To recover the original function, integrate its derivative to get an equation with an unknown constant C. Then, substitute the coordinates of the given point into this equation for x and y, solve for C, and rewrite the formula with this calculated value.

The topic explained

Why the derivative alone is not enough

When you differentiate a function, any constant term turns to zero. Because of this, infinitely many functions share the exact same derivative. For instance, the derivative of x squared plus 3 and the derivative of x squared minus 8 are both 2x.

Pinpointing one specific curve from that family requires an extra clue. Integrating the derivative yields an antiderivative plus an arbitrary constant, written as plus C. A single known point on the original curve acts as an anchor, allowing you to find the exact value of C.

The three-step method to solve for C

First, find the indefinite integral of the derivative. If f'(x) = 3x^2 - 4x + 1, integrating gives f(x) = x^3 - 2x^2 + x + C.

Second, plug the given point's coordinates into f(x). If the point is (2, 5), replace x with 2 and f(x) with 5 to get 5 = (2)^3 - 2(2)^2 + (2) + C, which simplifies to 5 = 2 + C.

Third, solve the linear equation for C and state the final formula. Here, C = 3, so the unique function is f(x) = x^3 - 2x^2 + x + 3.

Real-world context and common pitfalls

In physics, this same technique recovers position from velocity. If a car drives with velocity v(t) = 6t meters per second and starts 10 meters ahead of the origin at t = 0, the point is (0, 10). Integrating gives s(t) = 3t^2 + C, and substituting (0, 10) yields C = 10.

Watch out for signs when substituting negative x values into powers. Another common mistake is assuming C is always the y-value of the given point; this only happens when x equals zero and all other terms vanish.

Free printable worksheet

Finding a Function from Its Derivative and a Point

For 12th grade calculus students. Find the original function f(x) or position function s(t) using the derivative and given point.

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Personal AI lessons and practice, from preschool to university

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Finding a Function from Its Derivative and a PointName:

Find f(x) if f'(x) = 4 and the curve passes through (1, 6).
Find f(x) if f'(x) = 2x - 5 and f(0) = 4.
Find f(x) if f'(x) = 3x^2 + 2x and the graph passes through (2, 10).
Find f(x) if f'(x) = 6x^2 - 4x + 3 and f(1) = 8.
A particle moves with velocity v(t) = 12t^3 - 2t and position s(1) = 7. Find s(t).
Find f(x) if f'(x) = 1 / x^2 (for x > 0) and the curve passes through (2, 1).
Find f(x) given f'(x) = 3x^2 - 2, and the tangent line at x = 2 is y = 10x - 17.

Common questions

What is an initial condition in calculus?

An initial condition is a specified pair of values, usually denoting a starting state such as position at time zero. In pure math, it simply means any known point (x, y) on the curve that allows you to calculate the integration constant C.

Is the constant C always equal to the given y-value?

No. C equals y only if all terms involving x evaluate to zero when x is substituted. For example, if f'(x) = 2x + 3 and f(1) = 7, then 7 = 1^2 + 3(1) + C, which gives C = 3, not 7.

How do you find the function if given a tangent line instead of a point?

Evaluate the tangent line equation at the point of contact to get the point's y-coordinate. That gives you the coordinate pair (x, y). Then integrate the derivative and substitute this point to solve for C.

What topic comes after finding particular antiderivatives?

Once you can find specific functions using initial conditions, you move to definite integrals and the Fundamental Theorem of Calculus to compute exact areas under curves.

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