Find the intervals on which ƒ(𝓍) = ∫ₓ¹ (t―3) (t―6)¹¹ dt is in...

Question

Find the intervals on which ƒ(𝓍) = ∫ₓ¹ (t―3) (t―6)¹¹ dt is increasing and the intervals on which it is decreasing.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says let H of X be equal to the interval from X to 0 of the quantity of T + 2 in quantity multiplied by the quantity of T minus 5 in quantity rates to the 5th power DT by the intervals where H of X is increasing and where it is decreasing. So in order to find the intervals where HF X is increasing and decreasing, we need to find the critical points. And to find the critical points, we'll need to calculate the derivative with respect to X of HFX. So, if we look at taking the derivative with respect to X of H of X, and look at our definition for H of X, we would be taken derivative with respect to X, where X is a limit of integration. And we'd want to use the fundamental theorem of calculus in order to evaluate that derivative. However, in order to use the fundamental theorem of calculus, we need the X variable to be the upper limit of integration. But here, in our definition of H of X, it's the lower limit. So we're going to need to rewrite HF X to make X the upper limit. So we'll have 8 of X. And this is going to be equal to, and here we call that when we flip the limits of integration that we need to multiply our in Y minus size, so we'll have minus the integral from 0 to X of quantity of T + 2 in quantity multiplied by the quantity of T minus 5 in quantity raises to the 5th power dt. So now we can take a derivative with respect to X of H of X. So we'll call that H of X. And here we use the fundamental theorem of calculus to evaluate that derivative. And so by the fundamental theorem of calculus, H axes could be equal to minus the quantity of X plus 2 in quantity multiplied by the quantity of X minus 5, in quantity raised to the 5th power. So here we've just taken our integran and replaced with X. So now we need to find the critical points. And recall that the critical points occur whenever our first derivative is equal to zero or undefined. So H X here is a polynomial, so it is defined for all values of X. So the only critical points are going to be due to HXx being equal to 0. So we're going to to set minus quantity of X plus 2 in quantity multiplied by the quantity of X minus 5, in quantity raises to the 5th power equal to 0. And we need to solve this for X. So to do this, we'll just set each factor equal to 0 and solve for X. So, for the first factor, we're going to have X + 2 equal to 0. And solving this for X by subtracting 2 from both sides will have that X is equal to -2. Now, for the second factor, we'll have the quantity of X minus 5 in quantity raised to the 5th power, equal to 0, and taking the fifth root of both sides, we'll have that X minus 5 is equal to 0. And then adding 5 to both sides to solve for X, we'll see that X is equal to 5. So two critical points are going to be X minus 2 and X equals to 5. So, we're going to use our critical points to divide up the real number line into 3 different intervals and determine whether H of X is increasing, decrease or decreasing on these intervals. So the first interval that we're going to look at is going to be from minus infinity to -2. Now, on this interval, we have that X is going to be less than -2. And so to figure out the sign of H X, which is what's going to tell us whether H of X is increasing or decreasing, we just need to determine the sign of each of our factors in our equation for H X. So, when X is less than -2, the quantity of X + 2 is also going to be less than 0, so it's going to be negative. And if you look at the quantity of X minus 5 in quantity rates to the 5th power, this is also going to be less than 0, then it since we'll have the addition of two negative values, and that will give us a negative number, and when we raise a negative number to the 5th power, it remains negative. And so this means that each of X. is going to be negative as well, since we'll be multiplying two negative factors together, which will give us an overall positive value, but we have a minus sign in front of those two factors which will give us an overall negative value for H X. And recall that if the first derivative is negative on an interval, That means that the function HMX is going to be decreasing on that interval. Now, the next interval that we need to look at is going to be the interval from -2 to 5. And so on this interval, X is going to be between the values of minus 2 and 5. So if we look at the quantity of X + 2, We see that this quantity is going to be positive, so it's going to be greater than 0. And if we look at the quantity of X minus 5, in quantity raises to the power. We see that this is still going to be negative, so this is going to be less than 0. So that means that each prime of X. It's going to be Reader, then 0. So since H X is greater than 0 on this interval, meaning it's positive, that means that our function H of X is increasing. On this interval, And the final interval that we want to look at. is going to be the interval from 5 to infinity. And when this interval, X is going to be greater than 5. And so the quantity of X + 2 is going to be greater than 0. Since we're just adding two positive values together, and the quantity of X minus 5 in quantity rates to the 5th power. is also going to be greater than 0. So this means that each of X. It's going to be negative, so it's going to be less than 0. Since we're multiplying together two positive quantities, but out in front we have a minus sign, that makes the sign. Or HX overall negative, so it's gonna be less than 0. And since HX is negative on this interval, that means that our function H of X is going to be decreasing. On this interval. So, for our final answer, we see that the interval over which our function is increasing. It is going to be the open interval from -2 to 5. And it's our function is going to be decreasing. On the open from minus infinity. 2 minus 2 Union with the open interval from 5 to infinity. And so that is the answer that we were looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Find the intervals on which ƒ(𝓍) = ∫ₓ¹ (t―3) (t―6)¹¹ dt is increasing and the intervals on which it is decreasing.

Key Concepts

Fundamental Theorem of Calculus

Derivative and Increasing/Decreasing Functions

Critical Points

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