Question Statement
Using the area shown below in the figure, evaluate the integrals:
(i) ∫abf(x)dx
(ii) ∫bcf(x)dx
(iii) ∫cdf(x)dx
(iv) ∫acf(x)dx
(v) ∫bdf(x)dx
(vi) ∫adf(x)dx
Background and Explanation
This problem demonstrates the geometric interpretation of definite integrals as the area under a curve. When f(x)≥0 on an interval [a,b], the definite integral ∫abf(x)dx represents the area bounded by the curve y=f(x), the x-axis, and the vertical lines x=a and x=b. Additionally, definite integrals satisfy the additive property over adjacent intervals.
Solution
From the figure, it is clear that the area between a to b is A1=2. Since this region lies above the x-axis, the definite integral equals this geometric area:
∫abf(x)dx=A1=2
From the figure, it is clear that the area between b to c is A2=3. Therefore:
∫bcf(x)dx=A2=3
From the figure, it is clear that the area from c to d is A3=3.5. Therefore:
∫cdf(x)dx=A3=3.5
To find the integral over the combined interval [a,c], we apply the additive property of definite integrals by splitting the interval at point b:
∫acf(x)dx=∫abf(x)dx+∫bcf(x)dx=2+(3)( Using part (i) & (ii)) =2+3=5
Similarly, we split the interval [b,d] at point c and sum the areas from parts (ii) and (iii):
∫bdf(x)dx=∫bcf(x)dx+∫cdf(x)dx=3+3.5=6.5
For the entire interval [a,d], we decompose the integral into the sum of all three subintervals:
∫adf(x)dx=∫abf(x)dx+∫bcf(x)dx+∫cdf(x)dx=2+(3)+3.5=8.5
- Geometric interpretation of definite integrals: ∫abf(x)dx=Area under curve from a to b (when f(x)≥0)
- Additive property of definite integrals: ∫acf(x)dx=∫abf(x)dx+∫bcf(x)dx for a<b<c
- Interval decomposition: Breaking larger intervals into smaller adjacent subintervals to compute total area
Summary of Steps
- Identify the three individual areas from the figure: A1=2 (interval [a,b]), A2=3 (interval [b,c]), and A3=3.5 (interval [c,d]).
- For single-interval integrals (parts i-iii), assign the corresponding area value directly to the integral.
- For combined intervals (parts iv-vi), apply the additive property by splitting the integral at intermediate points (b or c).
- Substitute the known area values and sum them algebraically to obtain the final results (5, 6.5, and 8.5 respectively).