Think Twice

Calculus Level 3

True or False?

Define a function y = f ( x ) y=f(x) which is continuous, real-valued over an interval [ a , b ] [a,b] . So, the area under the curve f ( x ) f(x) within this interval is must be a b f ( x ) d x \displaystyle \int_{a}^{b} f(x)\, dx .

False True

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1 solution

Kushal Bose
Apr 11, 2017

The above statement is true if the curve does not change the sign.If it changes the sign then positive area(area above the X-axis) and negative area (Area below the X-axis) cancels and reduced to a value which is different from the actual area.

To find the actual area under the curve we need to find the roots of the curve then integrate part by part then add up the modulus values of them.Then total area can be found out.

For example : Let f ( x ) = cos x f(x)=\cos x .If we want to find area from [ 0 , π ] [0,\pi] then 0 π cos x d x = 0 \int_{0}^{\pi} \cos x \, dx=0 .But we know the actual area is not zero.Then the root is (within interval) π / 2 \pi/2 .So, we have to integrate 0 π / 2 cos x d x + π / 2 π cos x d x = 2 |\int_{0}^{\pi/2} \cos x \,dx| + |\int_{\pi/2}^{\pi} \cos x \,dx|=2 which is different from our integral

So, the above statement need not be always true.

Yes, you are completely correct

Md Zuhair - 4 years, 2 months ago

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This is one of the most common mistake.

Kushal Bose - 4 years, 2 months ago

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Yes, we all think only of the 1st quadrant.

Md Zuhair - 4 years, 2 months ago

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