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Let {eq}g\left( x \right) = \int\limits_{ - 12}^x {f\left( t \right)dt} {/eq}, where {eq}f {/eq} is the function whose graph is shown. Evaluate {eq}g\left( {-4} \right) {/eq}.

Question:

Let {eq}g\left( x \right) = \int\limits_{ - 12}^x {f\left( t \right)dt} {/eq}, where {eq}f {/eq} is the function whose graph is shown. Evaluate {eq}g\left( {-4} \right) {/eq}.

Integrability:


The integrability of a function is nothing but whether the given function is integrable in the given interval or not. If a function cannot be integrated into the given integral, the function is said to be a non-integrable function.

Answer and Explanation: 1

Given

  • Consider the given graph of {eq}f {/eq}. And, {eq}g\left( x \right) = \int\limits_{ - 12}^x {f\left( t \right)dt} {/eq} .


The objective is to evaluate the value of {eq}g\left( { - 4} \right) {/eq}.


Consider the given integral,

{eq}\begin{align*} g\left( x \right) &= \int\limits_{ - 12}^{ - 4} {f\left( t \right)} dt\\ g\left( { - 4} \right) &= \int\limits_{ - 12}^{ - 4} {f\left( { - 4} \right)} dx\\ &= \int\limits_{ - 12}^{ - 4} {12dx} \\ &= \left[ {12x} \right]_{ - 12}^{ - 4} \end{align*} {/eq}


Simplifying further, we get

{eq}\begin{align*} g\left( { - 4} \right) &= \left[ {12\left( { - 4} \right) - 12\left( { - 12} \right)} \right]\\ &= - 48 + 144\\ &= 96 \end{align*} {/eq}


Therefore, the required answer is,

{eq}g\left( { - 4} \right) = 96 {/eq}


Learn more about this topic:

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Integration Problems in Calculus: Solutions & Examples

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