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Calculate the approximation, f(x)=2+(x-2)^2 on the interval [-1,4].

Question:

Calculate the approximation, {eq}f(x)=2+(x-2)^2 {/eq}on the interval {eq}[-1,4] {/eq}.

Definite Integral:

Basically, integration is the opposite of differentiation. We know that definite integral can be expressed using the fundamental theorem of calculus, which expressed as {eq}\int_{a}^{b} f(x)dx=F(b)-F(a) {/eq}, where {eq}a {/eq} and {eq}b {/eq} are the intervals.

Answer and Explanation:

We can simply calculate the approximation of the given function using integration. In integrating the function, it is necessary to use the sum rule, defined as {eq}\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx {/eq}. Then, we have

{eq}\begin{align} \int_{-1}^42+\left(x-2\right)^2dx&= \int _{-1}^4x^2-4x+6dx \\ &= \int _{-1}^4 x^2dx-\int _{-1}^44xdx+\int _{-1}^46dx \\ &= \frac{x^3}{3}-4\left(\frac{x^2}{2}\right)+6x\big|_{-1}^4 \\ &= \frac{(4)^3}{3}-4\left(\frac{(4)^2}{2}\right)+6(4) -\left( \frac{(-1)^3}{3}-4\left(\frac{(-1)^2}{2}\right)+6(-1)\right)\\ &= \frac{40}{3}- \left(-\frac{25}{3}\right)\\ &= \frac{65}{3} \\ &= 21.67 \end{align} {/eq}


Learn more about this topic:

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Evaluating Definite Integrals Using the Fundamental Theorem

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2
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