How to find the concavity of an integral?

Question:

How to find the concavity of an integral?

Concavity of Function:

First we need to understand the concavity of a function:

Concave up/ Concave: {eq}y = f(x) {/eq} is concave up on an interval {eq}I {/eq} iff {eq}\displaystyle f''(x) > 0 {/eq} on the interval {eq}I {/eq}.

Concave down / Convex: {eq}y = f(x) {/eq} is concave down on an interval {eq}I {/eq} iff {eq}\displaystyle f''(x) < 0 {/eq} on the interval {eq}I {/eq}.

Answer and Explanation:


Suppose we have given an integral function {eq}\displaystyle H(x) = \int_{g(x)}^{h(x)} f(t) \ dt {/eq}.

First we will compute the first derivative of {eq}H(x) {/eq} using the Leibniz rule of the derivative.

The Leibniz rule of the derivative is defined as: {eq}\displaystyle \frac{d}{dx} \left [ \int_{g(x)}^{h(x)} f(t) \ dt \right ] = f(h(x)) \frac{d}{dx}[h(x)] - f(g(x)) \frac{d}{dx}[g(x)] {/eq}.

We can understand in the better way using an example. Compute the concavity of an integral {eq}\displaystyle f(x) = \int_0^x \frac{6}{t^2 + 7t + 18} dt {/eq}


Take the derivative of {eq}f(x) {/eq} with respect to x

$$\begin{align*} \displaystyle f'(x) &= \frac{d}{dx} \left [ \int_0^x \frac{6}{t^2 + 7 t + 18} \right ] \\ &= \frac{6}{x^2 + 7 x + 18} \frac{d}{dx}[x] - \frac{6}{0^2 + 7 \times 0 + 18} \frac{d}{dx}[0] \\ &= \frac{6}{x^2 + 7 x + 18} - 0 &\text{(Differentiating using the power formula } \frac{d}{dx}[x^n] = n x^{n - 1} \text{)}\\ f'(x) &= \frac{6}{x^2 + 7 x + 18} \\ f''(x) &= \frac{d}{dx}\left [ \frac{6}{x^2 + 7 x + 18} \right ] &\text{(Taking the derivative of } f'(x) \text{ with respect to x)}\\ &= \frac{(x^2 + 7 x + 18) \frac{d}{dx}[6] - 6 \frac{d}{dx}[x^2 + 7 x + 18] }{(x^2 + 7 x + 18)^2} &\text{(Applying the quotient rule of the derivative)}\\ &= \frac{(x^2 + 7 x + 18) \times 0 - 6 (2 x + 7 + 0) }{(x^2 + 7 x + 18)^2} &\text{(Differentiating using the power formula } \frac{d}{dx}[x^n] = n x^{n - 1} \text{)}\\ f''(x) &= \frac{- 6 (2 x + 7)}{(x^2 + 7 x + 18)^2}\\ \end{align*} $$


As we know {eq}f''(x) > 0 {/eq} for the curve to be concave up.

$$\begin{align*} \displaystyle f''(x) &> 0\\ \frac{- 6 (2 x + 7)}{(x^2 + 7 x + 18)^2} &> 0 &\text{(Plugging in the value of } f''(x) \text{)}\\ - 6 (2 x + 7) &> 0 &\text{(Multiplying both sides by } (x^2 + 7 x + 18)^2 \text{)}\\ 2 x + 7 &< 0 &\text{(Dividing both sides by - 6)}\\ 2 x &< - 7 \\ x &< - \frac{7}{2} &\text{(Dividing both sides by 2)}\\ \end{align*} $$


Thus, the curve will be concave up on the interval {eq}\displaystyle \left (- \infty, - \frac{7}{2} \right ) {/eq} and concave down on the interval {eq}\displaystyle \left (- \frac{7}{2}, \infty \right ) {/eq}.


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