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Evaluate the definite integral using linearly and subdivision together with the following...

Question:

Evaluate the definite integral using linearly and subdivision together with the following results. {eq}\int\limits_{-1}^2 x^2 d x = 3; \int\limits_{-1}^0 x^2 d x = \frac{1}{3}; \int\limits_{-1}^{2} x d x = \frac{3}{2}; \int\limits_0^2 x d x = 2 {/eq} Find {eq}\int\limits_{-1}^2 (x^2 + x) d x = {/eq}

Integration:


Integration is reverse of differentiation.

We will use rule:

{eq}\displaystyle \int \:\left[f\left(x\right)\pm g\left(x\right)\right]dx=\int \:f\left(x\right)dx\pm \int \:g\left(x\right)dx {/eq}

Answer and Explanation:


Given integral is

{eq}\displaystyle I=\int _{-1}^2\left(x^2\:+\:x\right)\:dx\:\\ \displaystyle I=\int _{-1}^2x^2\:dx+\int _{-1}^2\:x\:dx\: {/eq}

Given values are

{eq}\displaystyle \int\limits_{-1}^2 x^2 d x = 3; \int\limits_{-1}^0 x^2 d x = \frac{1}{3}; \int\limits_{-1}^{2} x d x = \frac{3}{2}; \int\limits_0^2 x d x = 2 {/eq}

{eq}\displaystyle \therefore\:I=3+\frac{3}{2}\\ \displaystyle I=\frac{9}{2} {/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
1.9K

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