Video: Finding the General Antiderivative of a Function

What is the antiderivative 𝐹 of 𝑓(π‘₯) = βˆ’5 + (1 + π‘₯Β²)⁻¹ that satisfies 𝐹(1) = 0?

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Video Transcript

What is the antiderivative capital 𝐹 of 𝑓 of π‘₯ equals negative five plus one plus π‘₯ squared to the power of negative one that satisfies capital 𝐹 of one equals zero?

Here, we’ve been given a rather nasty looking function and asked to work out its antiderivative. That is, the function that, when differentiated, gives us negative five plus one plus π‘₯ squared to the power of negative one. Now, if we rewrite one plus π‘₯ squared to the power of negative one as one over one plus π‘₯ squared, we can quote the general result for the derivative of the inverse tan of π‘Žπ‘₯. It’s π‘Ž over one plus π‘Ž of π‘₯ all squared for real constants π‘Ž. Then, if we let π‘Ž be equal to one, we find that the derivative of the inverse tan of π‘₯ is one over one plus π‘₯ squared. Well, we actually have that expression in our function. And that tells us that the antiderivative of one over one plus π‘₯ squared must be the inverse tan of π‘₯. So what about the antiderivative of negative five?

Well, the antiderivative of negative five is negative five π‘₯. So we find that an antiderivative capital 𝐹 could be negative five π‘₯ plus inverse tan of π‘₯. Remember though, we need to include some constant. And this is because when we differentiate a constant, we end up with zero. Luckily, we can work out the value of this constant by using the fact that capital 𝐹 of one is equal to zero. Let’s substitute π‘₯ equals one into our expression for the antiderivative. When we do, we obtain zero to be equal to negative five times one plus the inverse tan of one plus 𝑐. Negative five times one is negative five. And the inverse tan of one is πœ‹ by four. We solve this equation by adding five to both sides and then subtracting πœ‹ by four. And we see that 𝑐 is equal to five minus πœ‹ by four.

And so we see the antiderivative capital 𝐹 is defined by the function negative five π‘₯ plus the inverse tan of π‘₯ minus πœ‹ by four plus five.

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