Question Statement

Find: ${\int }_1^{4}f(x)dx$

given that: ${\int }_1^{2}f(x)dx=1\text{and}{\int }_2^{4}f(x)dx=2$

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Background and Explanation

This problem relies on the additive property of definite integrals, which states that the integral over a larger interval can be decomposed into the sum of integrals over adjacent subintervals. This property is essential when we know the behavior of a function over specific segments but need to calculate the total accumulation over the combined range.

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Solution

The interval $[1,4]$ can be split at the point $x=2$ into two adjacent subintervals: $[1,2]$ and $[2,4]$. Since we are given the values of the integral over each of these subintervals, we can apply the additive property of definite integrals to find the total integral from $1$ to $4$.

Using the property that ${\int }_a^{c}f(x)dx={\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx$ where $a=1$, $b=2$, and $c=4$:

$\begin{array}{rl}{\int }_1^{4}f(x)dx& ={\int }_1^{2}f(x)dx+{\int }_2^{4}f(x)dx\\ & =1+2\\ & =3\end{array}$

Therefore, the value of the definite integral is $3$.

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Key Formulas or Methods Used

• Additive Property of Definite Integrals: ${\int }_a^{c}f(x)dx={\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx$ (where $a<b<c$)
• Interval Decomposition: Breaking a larger integration interval into contiguous subintervals at known boundary points

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Summary of Steps

1. Identify that the interval $[1,4]$ can be split at $x=2$ into the subintervals $[1,2]$ and $[2,4]$
2. Apply the additive property: ${\int }_1^{4}f(x)dx={\int }_1^{2}f(x)dx+{\int }_2^{4}f(x)dx$
3. Substitute the given values: $1+2$
4. Calculate the final result: $3$