# Video: Finding the Expression of a Function given the Expression of Its First Derivative and the Value of the Function at a Point Using Integration

Determine the function 𝑓(𝑡) such that 𝑓′(𝑡) = −2/(3(𝑡² + 1)), and 𝑓(1) = 0.

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### Video Transcript

Determine the function 𝑓 of 𝑡 such that 𝑓 prime of 𝑡 is equal to negative two divided by three times 𝑡 squared plus one and 𝑓 evaluated at one is equal to zero.

The question wants us to determine the function 𝑓 of 𝑡. The information we’re given is that the derivative of 𝑓 of 𝑡 with respect to 𝑡 is equal to negative two divided by three times 𝑡 squared plus one and 𝑓 evaluated at one is equal to zero. This is a differential equation. In fact, it’s a simple differential equation because we’re just given 𝑓 prime of 𝑡 in terms of 𝑡. So to solve this, we’re going to integrate both sides of this equation with respect to 𝑡. This will give us our function 𝑓 of 𝑡 up to a constant of integration. This gives us the integral of 𝑓 prime of 𝑡 with respect to 𝑡 is equal to the integral of negative two divided by three times 𝑡 squared plus one with respect to 𝑡.

Remember, integration and differentiation are opposite processes. So, the integral of 𝑓 prime of 𝑡 with respect to 𝑡 is 𝑓 of 𝑡 up to a constant of integration. So to find 𝑓 of 𝑡, we just need to evaluate our integral. Before we evaluate our integral, we’ll take out our constant factor of negative two divided by three. This gives us negative two over three times the integral of one divided by 𝑡 squared plus one with respect to 𝑡. But this integral is just one of our standard integrals involving inverse trigonometric functions.

We know the integral of one divided by 𝑥 squared plus one with respect to 𝑥 is equal to the inverse tan of 𝑥 plus the constant of integration 𝐶. So, we can just apply this to evaluate our integral. We get negative two over three times the inverse tan of 𝑡 plus 𝐶. So, this is the general solution to our differential equation. Remember, we want the solution where 𝑓 evaluated at one is equal to zero. So, we’ll substitute this information into our general solution to find the value of 𝐶.

This gives us zero is equal to negative two-thirds times the inverse tan of one plus 𝐶. We can simplify this equation. We can divide both sides through by negative two-thirds. Next, we can evaluate the inverse tan of one; it’s equal to 𝜋 by four. Finally, we subtract 𝜋 by four from both sides of this equation. We see that 𝐶 is equal to negative 𝜋 by four. All we need to do now is use this value of 𝐶 in our general solution.

Doing this, we get 𝑓 of 𝑡 is equal to negative two over three times the inverse tan of 𝑡 minus 𝜋 over four. And we could leave our answer like this. However, we’ll distribute negative two-thirds over our parentheses. This gives us negative two times the inverse tan of 𝑡 divided by three plus 𝜋 by six. And this gives us our final answer.

Therefore, we’ve shown if 𝑓 prime of 𝑡 is equal to negative two over three times 𝑡 squared plus one and 𝑓 of one is equal to zero. Then 𝑓 of 𝑡 is equal to negative two times the inverse tan of 𝑡 divided by three minus 𝜋 by six.