Question Video: Area Under a Curve as an Integral | Nagwa Question Video: Area Under a Curve as an Integral | Nagwa

# Question Video: Area Under a Curve as an Integral Mathematics • Third Year of Secondary School

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The curve shown is π¦ = 1/π₯. What is the area of the shaded region? Give an exact answer.

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### Video Transcript

The curve shown is π¦ equals one over π₯. What is the area of the shaded region? Give an exact answer.

We recall first of all that in order to find the area of a region bounded by a curve, we can use integration. The exact area of the region bounded by the curve π¦ equals π of π₯, the π₯-axis, and the two vertical lines π₯ equals π and π₯ equals π is given by the definite integral from π to π of π of π₯ with respect to π₯. If the area we were looking to calculate was below the π₯-axis, we would need to take the absolute value of this integral as the integral itself would give a negative result. But we can see from the figure that the area weβre looking to find is above the π₯-axis. So this isnβt a concern here. We need to determine then the function π of π₯ and the values of π and π in this problem.

π of π₯ is given in the question. Itβs the function one over π₯. For the values of π and π, we can use the given figure. We see that this region is bounded on the left by the line π₯ equals one. And so the lower limit for our integral is one. The region is bounded on the right by the line π₯ equals three. And so the upper limit for our integral is three. We have then that the area is equal to the definite integral from one to three of one over π₯ with respect to π₯.

We then recall that the integral of one over π₯ with respect to π₯ is the natural logarithm of the absolute value of π₯ plus a constant of integration if performing an indefinite integral. Now, here, the values of π₯ for our limit are both positive. So we donβt actually need to include the absolute value signs around that π₯ inside the natural logarithm. We can say then that this area is equal to the natural logarithm of π₯ evaluated between one and three. Substituting the limits gives the natural logarithm of three minus the natural logarithm of one. And at this point, we can recall that the natural logarithm of one is just zero. We are asked to give an exact answer. So weβll leave our answer in terms of a natural logarithm rather than evaluating on a calculator and rounding.

Using integration then, we found that the exact area of the shaded region is the natural logarithm of three square units.

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