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 Functionβs Average Value over It Mathematics • Higher Education

The average value of π(π₯) = β6π₯Β² + 6π₯ β 1 on the interval [0, π] is 0. Find all possible values of π.

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### Video Transcript

The average value of π of π₯ equals negative six π₯ squared plus six π₯ minus one on the closed interval zero to π is zero. Find all possible values of π.

Remember, the formula for the average value of a function π over a closed interval π to π is one over π minus π times the integral evaluated between π and π of π of π₯ with respect to π₯. We can see that π of π₯ is equal to negative six π₯ squared plus six π₯ minus one. And we see that π is equal to zero, and we donβt know the value of π. So we begin by substituting what we know about the average value of our function into the formula.

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

Letβs look to evaluate our integral. The integral of negative six π₯ squared is negative six π₯ cubed over three. The integral of six π₯ is six π₯ squared over two. And the integral of negative one is negative π₯. Letβs substitute π and zero into this expression. We see that zero is equal to one over π times negative two π cubed plus three π squared minus π. And then we divide through by π. And so we obtain that negative two π squared plus three π minus one is equal to zero. And we see we have a quadratic equation.

So letβs multiply through by negative one. And then weβll factor the expression two π squared minus three π plus one. When we do, we see that two π minus one times π minus one must be equal to zero. And for this statement to be true, either two π minus one must be equal to zero or π minus one must be equal to zero. And we solve for π. And we see that π must be equal to one-half or one.

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