Question Video: Finding an Interval given the Function’s Average Value over It | Nagwa Question Video: Finding an Interval given the Function’s Average Value over It | Nagwa

Question Video: Finding an Interval given the Function’s Average Value over It Mathematics

The average value of 𝑓(𝑥) = −6𝑥² + 6𝑥 − 1 on the interval [0, 𝑏] is 0. Find all possible values of 𝑏.

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Video Transcript

The average value of 𝑓 of 𝑥 equals negative six 𝑥 squared plus six 𝑥 minus one on the closed interval zero to 𝑏 is zero. Find all possible values of 𝑏.

Remember, the formula for the average value of a function 𝑓 over a closed interval 𝑎 to 𝑏 is one over 𝑏 minus 𝑎 times the integral evaluated between 𝑎 and 𝑏 of 𝑓 of 𝑥 with respect to 𝑥. We can see that 𝑓 of 𝑥 is equal to negative six 𝑥 squared plus six 𝑥 minus one. And we see that 𝑎 is equal to zero, and we don’t know the value of 𝑏. So we begin by substituting what we know about the average value of our function into the formula.

We get one over 𝑏 minus zero, which is of course one over 𝑏. And we multiply that by the definite integral of negative six 𝑥 squared plus six 𝑥 minus one evaluated between zero and 𝑏. We do, however, know that the average value of our function is equal to zero. So we can set this equal to zero.

Let’s look to evaluate our integral. The integral of negative six 𝑥 squared is negative six 𝑥 cubed over three. The integral of six 𝑥 is six 𝑥 squared over two. And the integral of negative one is negative 𝑥. Let’s substitute 𝑏 and zero into this expression. We see that zero is equal to one over 𝑏 times negative two 𝑏 cubed plus three 𝑏 squared minus 𝑏. And then we divide through by 𝑏. And so we obtain that negative two 𝑏 squared plus three 𝑏 minus one is equal to zero. And we see we have a quadratic equation.

So let’s multiply through by negative one. And then we’ll factor the expression two 𝑏 squared minus three 𝑏 plus one. When we do, we see that two 𝑏 minus one times 𝑏 minus one must be equal to zero. And for this statement to be true, either two 𝑏 minus one must be equal to zero or 𝑏 minus one must be equal to zero. And we solve for 𝑏. And we see that 𝑏 must be equal to one-half or one.

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