What is the area of the region in the first quadrant bounded by the graph of y = e^x/2 and the line x = 2?

Solution:

The following diagram represents the graph of y = e^x/2 and the line x = 2.

y = e^x/2 and the line x = 2

The area bounded by x = 2 and curve y = \(e^{x/2}\) in the first quadrant can be written as the sum of two regions i.e. Area ABC and Area ACOE.

The area ABC can be written as an integral of the function y = \(e^{x/2}\) over the limits of x = 0 to x = 2.

Area ABC = \(\int_{0}^{2}e^{x/2}dx\)

= (1/2)\([e^{x/2}]_{0}^{2}\)

= (1/2)\([e^{(2/2)} - e^{0/2}]\)

= (1/2)\([e^{1} - e^{0}]\)

= (1/2)[ 2.718 - 1]

= (1/2)[1.718]

= 0.859 unit²

Area AEOC is simply : The area of the rectangle:

Area AEOC = OE × AE = 2 × 1 = 2

The total area bounded by the curve y = \(e^{x/2}\) and x = 2 is given by

Total Area = Area ACB + Area AEOC

= 0.859 + 2

= 2.859 unit².

What is the area of the region in the first quadrant bounded by the graph of y = e^x/2 and the line x = 2?

Summary:

The area of the region in the first quadrant bounded by the graph of y = e^x/2 and the line x = 2 is 2.859 unit².