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Math · Calculus I · Worked example

Solve an initial-value problem

Find f(x) if f′(x) = 3x² + 2 and f(1) = 5.

f⁡′(x)=3x2+2,f⁡(1)=5

Antidifferentiate

Every antiderivative of 3x² + 2 has the form x³ + 2x + C.

f⁡(x)=x3+2⁢x+C

Use the condition

Substitute x = 1 and set the result equal to 5: 1 + 2 + C = 5, so C = 2.

1+2+C=5⟹C=2

Write the function and check it

f(x) = x³ + 2x + 2. Its derivative is 3x² + 2, and f(1) = 1 + 2 + 2 = 5, so both conditions hold.

f⁡(x)=x3+2⁢x+2

Result

f(x) = x³ + 2x + 2.

Your turn

A ball is thrown upward at 20 m/s from a height of 2 m, so its velocity is v(t) = 20 − 9.8t m/s. Find its height h(t).

Show the answer and explanation

h(t) = 2 + 20t − 4.9t² meters.

Height is an antiderivative of velocity: h(t) = 20t − 4.9t² + C. The starting height h(0) = 2 gives C = 2. Differentiating gives back 20 − 9.8t.

h⁢(t)=2+20⁢t−4.9t2h′(t)=20−9.8⁢th⁢(0)=2

Keep exploring

In Math, change the last row to f(1) = 6 and the box marks it wrong. Only C = 2 fits: every other constant shifts the curve up or down, off the point (1, 5).

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