Question

Evaluate the following integrals by interpreting each in terms of areas. (a) int_{0}^{5} sqrt{25-x^{2}} dx (b) int_{0}^{4}(x-1) dx Solution (a) Since f(x) = sqrt{25-x^{2}} geq 0, we can interpret this integral as the area under the curve y = sqrt{25-x^{2}} from 0 to 5. But, because y^2 = 25 - x^2, we get x^2 + y^2 = 25, which shows that the graph of f is the quarter-circle with radius 5 in the figure below. Therefore, int_{0}^{5} sqrt{25-x^{2}} dx = frac{1}{4}pi(5)^2 = frac{25pi}{4}. (b) The graph of y = x - 1 is the line with slope 1 shown in the following figure. We compute the integral as the difference of the areas of the two triangles. int_{0}^{4}(x-1) dx = A_1 - A_2 = square - 0.5 = square.

          Evaluate the following integrals by interpreting each in terms of areas.
(a) int_{0}^{5} sqrt{25-x^{2}} dx
(b) int_{0}^{4}(x-1) dx
Solution
(a) Since f(x) = sqrt{25-x^{2}} geq 0, we can interpret this integral as the area under the curve y = sqrt{25-x^{2}} from 0 to 5. But, because y^2 = 25 - x^2, we get x^2 + y^2 = 25, which shows that the graph of f is the quarter-circle with radius 5 in the figure below.
Therefore, int_{0}^{5} sqrt{25-x^{2}} dx = frac{1}{4}pi(5)^2 = frac{25pi}{4}.
(b) The graph of y = x - 1 is the line with slope 1 shown in the following figure.
We compute the integral as the difference of the areas of the two triangles.
int_{0}^{4}(x-1) dx = A_1 - A_2 = square - 0.5 = square.

$Evaluate the following integrals by interpreting each in terms of areas. (a) int0^5 sqrt25-x^2 dx (b) int0^4(x-1) dx Solution (a) Since f(x) = sqrt25-x^2 geq 0, we can interpret this integral as the area under the curve y = sqrt25-x^2 from 0 to 5. But, because y^2 = 25 - x^2, we get x^2 + y^2 = 25, which shows that the graph of f is the quarter-circle with radius 5 in the figure below. Therefore, int0^5 sqrt25-x^2 dx = frac14pi(5)^2 = frac25pi4. (b) The graph of y = x - 1 is the line with slope 1 shown in the following figure. We compute the integral as the difference of the areas of the two triangles. int0^4(x-1) dx = A1 - A2 = square - 0.5 = square.$

Added by Charles C.

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