Question

Find the area under the curve of the function
between the limits 0 and 4.

Answer 1

`\textsf{C.} \quad (\pi)/(16)`

Explanation:

To find the area under the curve of a function f(x) between specified limits, we can use definite integration.

Definite integration

The definite integral of f(x) with respect to x between the limits x = a and x = b is:

`\boxed{\displaystyle \int^b_a \text{f}(x)\; \text{d}x=\left[\text{g}(x)\right]^b_a=\text{g}(b)-\text{g}(a)}`

where a is the lower limit and b is the upper limit.

Given function:

`f(x)=(1)/(x^2+16)`

First we need to determine if the function f(x) crosses the x-axis between the given limits of x = 0 and x = 4. (If any part of the curve between these limits is below the x-axis, we will need to integrate each part separately).

A function crosses the x-axis when y = 0. As the function is a rational function where the denominator is greater than or equal to 16, then y > 0. Therefore, the curve is always above the x-axis, and is continuous between the given limits.

Therefore, the definite integral to evaluate is:

`\displaystyle \int^4_0 (1)/(x^2+16)\; \text{d}x`

To evaluate the integral, use the method of substitution.

`\textsf{Let}\;\; u = (x)/(4) \implies x=4u`

Differentiate u with respect to x to find du/dx:

`u=(x)/(4) \implies \frac{\text{d}u}{\text{d}x}=(1)/(4)`

Rearrange to isolate dx:

`\text{d}x=4\;\text{d}u`

Use the substitution to change the limits of the integral from x-values to u-values:

`x=0 \implies u=(0)/(4)=0`

`x=4 \implies u=(4)/(4)=1`

As x = 4u, then function f(u) is:

`(1)/(x^2+16)=(1)/((4u)^2+16)=(1)/(16u^2+16)=(1)/(16(u^2+1))`

Substitute everything back into the original integral to create a definite integral in terms of u:

`\begin{aligned}\displaystyle \int^4_0 (1)/(x^2+16)\; \text{d}x&=\int^(1)_(0) (1)/(16(u^2+1)) \cdot 4\;\text{d}u\\\\&=\int^1_0 (4)/(16(u^2+1)) \;\text{d}u\\\\&=\int^1_0 (1)/(4(u^2+1)) \;\text{d}u\\\\&=(1)/(4)\int^1_0 (1)/(u^2+1) \;\text{d}u\end{aligned}`

Use the following integration formula to evaluate the integral:

`\boxed{\displaystyle \int (1)/(1+u^2)\; \text{d}u=\arctan u + \text{C}}`

Therefore:

`\begin{aligned}\displaystyle \int^4_0 (1)/(x^2+16)\; \text{d}x&=(1)/(4)\int^1_0 (1)/(u^2+1) \;\text{d}u\\\\&=(1)/(4)\left[\vphantom{\frac12}\arctan u\right]^1_0\\\\&=(1)/(4)\arctan (1)-(1)/(4) \arctan(0)\\\\&=(1)/(4)\cdot (\pi)/(4)-(1)/(4) \cdot 0\\\\&=(\pi)/(16)\end{aligned}`

So the area under the curve of the function f(x) between the limits 0 and 4 is π/16.