Mathematics

Mathematics

Problem

Two differentiable functions hh and kk intersect at several points on the interval [c,d][c,d]. A student claims that ∫cd∣h(x)−k(x)∣ dx\int_{c}^{d}\lvert h(x)-k(x)\rvert\,dx represents the total area of the region enclosed by the two curves without requiring the student to find those intersection points. Which of the following statements best justifies the student’s claim? a) Because one may first compute ∫cd(h(x)−k(x)) dx\displaystyle\int_{c}^{d}\bigl(h(x)-k(x)\bigr)\,dx and then take the absolute value of that single number, yielding the total area between the curves. b) Because the absolute value forces the integrand to be non-negative, any portion where h(x)<k(x)h(x)<k(x) contributes positively to the integral instead of canceling out. c) Because ∣h(x)−k(x)∣=h(x)−k(x)\lvert h(x)-k(x)\rvert=h(x)-k(x) on the entire interval [c,d][c,d], the integrand is always non-negative. d) Because definite integrals are automatically non-negative, the sign of h(x)−k(x)h(x)-k(x) does not affect the value of the integral.
Intuitive explanation

The absolute value automatically handles the sign changes that occur whenever the curves cross—it flips the integrand from h(x)−k(x)h(x)-k(x) to k(x)−h(x)k(x)-h(x) (or vice versa) at each intersection point, ensuring every enclosed region contributes positively to the sum. This means you're adding up all the vertical distances between the curves across the entire interval, regardless of which function is on top, without needing to split the integral at each crossing point manually.

Intuitive explanation

The absolute value automatically handles the sign changes that occur whenever the curves cross—it flips the integrand from h(x)−k(x)h(x)-k(x) to k(x)−h(x)k(x)-h(x) (or vice versa) at each intersection point, ensuring every enclosed region contributes positively to the sum. This means you're adding up all the vertical distances between the curves across the entire interval, regardless of which function is on top, without needing to split the integral at each crossing point manually.

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