# Consider the function f(x) = 3 + \int^{1+x^2}_2 5/2(1 + t^2) dt. (a) Find the linearization of...

## Question:

Consider the function f(x) = 3 +{eq}\int^{1+x^2}_2 5/2(1 + t^2) {/eq} dt.

(a) Find the linearization of f(x) at a = 1.

(b) Use the linearization in (a) to approximate {eq}\int^{1+(1.1)^2}_2 5/2(1 + t^2) {/eq}dt.

## Linearization:

When functions are extremely difficult to evaluate, the process of linearization helps us approximate the value of the function. The formula to perform linearization is given as {eq}f(x) = f(x_0) + h \cdot f'(x_{0}) {/eq}, where {eq}f(x)_0 {/eq}, {eq}h = x - x_0 {/eq}, and {eq}f'(x_{0}) {/eq}, are provided in order to find the {eq}f(x) {/eq} value.

Given: {eq}f(x) = 3+\int^{1+x^2}_2 \frac{5}{2}(1+t^2) dt {/eq}:

a. The point of reference will be at {eq}x = a = 1 {/eq}:

The step size, accordingly, will be {eq}h = x-1 {/eq}

The derivative of {eq}f(x) {/eq} in general is

{eq}\begin{align*} f'(x) &= \frac{d}{dx}(f(x)) \\ &= \frac{d}{dx}(3+\int^{1+x^2}_2 \frac{5}{2}(1+t^2) dt) \\ &= \frac{d}{dx}(3+\int^{1+x^2}_2 \frac{5}{2}+\frac{5t^2}{2} dt) \\ &= \frac{d}{dx}(3+\left[\frac{5}{2}t+\frac{5t^{2+1}}{2(2+1)}\right]^{1+x^2}_2) \\ &= \frac{d}{dx}(3+\left[\frac{5}{2}t+\frac{5t^{3}}{2(3)}\right]^{1+x^2}_2) \\ &= \frac{d}{dx}(3+\left[\frac{5}{2}t+\frac{5t^{3}}{6}\right]^{1+x^2}_2) \\ &= \frac{d}{dx}(3+(\frac{5}{2}(1+x^2)+\frac{5(1+x^2)^{3}}{6})-(\frac{5}{2}(2)+\frac{5(2)^{3}}{6})) \\ &= \frac{d}{dx}(3+(\frac{5}{2}(1+x^2)+\frac{5(1+x^2)^{3}}{6})-(\frac{5}{2}(2)+\frac{5(2)^{3}}{6})) \\ &= \frac{d}{dx}(3+((\frac{5}{2}+\frac{5x^2}{2})+\frac{5(1+x^2)^{3}}{6})-(\frac{10}{2}+\frac{5(8)}{6})) \\ &= \frac{d}{dx}(3+\frac{5}{2}+\frac{5x^2}{2}+\frac{5(1+x^2)^{3}}{6}-5+\frac{40}{6}) \\ &= \frac{d}{dx}(\frac{18}{6}+\frac{15}{6}+\frac{5x^2}{2}+\frac{5(1+x^2)^{3}}{6}-\frac{30}{6}+\frac{40}{6}) \\ &= \frac{d}{dx}(\frac{5x^2}{2}+\frac{5(1+x^2)^{3}}{6}+\frac{43}{6}) \\ &= \frac{2\cdot 5x^{2-1}}{2}+\frac{3\cdot (1+x^2)^{3-1}\cdot (2\cdot x)}{6}+0 \\ &= 5x+\frac{6x(1+x^2)^2}{6} \\ &= 5x+x(1+x^2)^2 \\ \end{align*} {/eq}

Now all the components to compute {eq}f(x) {/eq} are found, so simply apply the linear approximation formula to find {eq}L(x) {/eq}.

{eq}\begin{align*} L(x) = f(1)+h\cdot f'(1) &= (3+\int^{1+(1)^2}_2 \frac{5}{2}(1+t^2))+x(5(1)+(1)(1+(1)^2)^2) \\ &= (3+\int^{1+1}_2 \frac{5}{2}(1+t^2))+x(5+1(1+1)^2) \\ &= (3+\int^{2}_2 \frac{5}{2}(1+t^2))+x(5+1(2)^2) \\ &= (3+0)+x(5+1(4)) \text{ [Since integral bounds are the same]}\\ &= (3)+x(5+4) \\ &= 9x+3 \\ \end{align*} {/eq}

b. Find: {eq}\int^{1+(1.1)^2}_2 \frac{5}{2}(1+t^2) dt {/eq}

By inspection, this is the same as {eq}f(1.1)-3 {/eq} since the most of the expression resembles {eq}f(x) {/eq} when {eq}x = 1.1 {/eq}.

Therefore, using the linearization in part a, the formula {eq}L(1.1)-3 {/eq} will find the approximate value to {eq}\int^{1+(1.1)^2}_2 \frac{5}{2}(1+t^2) dt {/eq}

{eq}\begin{align*} \int^{1+(1.1)^2}_2 \frac{5}{2}(1+t^2) dt &= L(1.1)-3 \\ &= (9(1.1)+3)-3 \\ &= 9.9+3-3 \\ &= 9.9 \\ \end{align*} {/eq}