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Use the concept of the definite integral to find the total area between the graph off(x) and thex-axis, by taking the limit of the associated right Riemann sum. Write the exact answer. Do not round. (Hint: Extra care isneeded on those intervals wheref(x) < 0. Remember that the definite integral represents a signed area.)f(x) = 3x + 3 on [0, 2]

Use The Concept Of The Definite Integral To Find The Total Area Between The Graph Offx And Thexaxis By Taking The Limit Of The Associated Right Riemann Sum Writ class=

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ANSWER

15 square unit.

EXPLANATION

Step 1:

Recall that the right endpoint Riemann sum for

[tex]\begin{gathered} ^{}\int ^b_af(x)dx\text{ is given by } \\ \frac{b-a}{n}\sum ^n_{k\mathop=1}f(a+\frac{b-a}{n}k) \end{gathered}[/tex]

Step 2:

Note, if f(x) is continuous, then:

[tex]\lim _{n\to\infty}\frac{b-a}{n}\sum ^n_{k\mathop{=}1}f(a+\frac{b-a}{n}k)\text{ = }^{}\int ^b_af(x)dx\text{ }[/tex]

Step 3:

Now, applying the limit of the Reimann sums to evaluate the integral:

[tex]^{}\int ^2_0(3x+3)dx\text{ }[/tex]

Please, carefully check my working:

Hence, using the concept of the definite integral, the total area between the graph of f(x) and the x-axis by taking the limit of the associated right Riemann sum is 15 square unit.

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