Question

Express with the aid of an integral the area of a figure bounded by : (i) The coordinate axes, the straight line \(\mathrm{x}=3\) and the parabola \(\mathrm{y}=\mathrm{x}^{2}+1\). (ii) The \(x\)-axis, the straight lines \(x=a, x=b\) and the curve \(y=e^{x}+2(b>a)\).

Short answer

Question: Determine the area of each figure enclosed by the given boundaries. Figure 1: Bounded by the coordinate axes, the line x=3, and the parabola y=x^2+1. Figure 2: Bounded by the x-axis, the lines x=a and x=b, and the curve y=e^x+2, with b>a. Answer: The area of Figure 1 is 12 square units. The area of Figure 2 is given by the expression: e^b-e^a+2(b-a).

Step-by-step solution

Set up the first integral

For the first figure, we want to find the area enclosed by the coordinate axes, the line x=3, and the parabola y=x^2+1. Notice that since the parabola will always be above the x-axis, we can integrate the parabola function with respect to x from 0 to 3 and find the area under the curve. The parabola function is given by y=x^2+1. Therefore, we set up the integral like this: $$ \int_{0}^{3} (x^2+1) dx $$

Evaluate the first integral

Now we need to evaluate the integral: $$ \int_{0}^{3} (x^2+1) dx = \left[ \frac{x^3}{3}+x \right]_0^3 $$ Plug in the limits of integration: $$ \left[ \frac{(3)^3}{3}+(3) \right] - \left[ \frac{(0)^3}{3}+(0) \right] = \frac{27}{3}+3=12 $$ The area of the first figure is 12 square units.

Set up the second integral

For the second figure, we want to find the area enclosed by the x-axis, the lines x=a and x=b, and the curve y=e^x+2, with b>a. We will integrate the curve function with respect to x from a to b. The curve function is given by y=e^x+2. Therefore, we set up the integral like this: $$ \int_{a}^{b} (e^x+2) dx $$

Evaluate the second integral

Now we need to evaluate the integral: $$ \int_{a}^{b} (e^x+2) dx = \left[ e^x + 2x \right]_a^b $$ Plug in the limits of integration: $$ \left[ e^b + 2b \right] - \left[ e^a + 2a \right] = e^b-e^a+2(b-a) $$ The area of the second figure is given by the expression: $$ e^b-e^a+2(b-a) $$