If area of shaded portion in figure A is A 1 and that of shaded portion in figure B is A 2- Then- A- A 1- A 2B- A 1- A 2C- A 1- A 2D- Insufficient inf ormation

The correct option is C A_1 = A_2As we already know the area under the curve is given by the integral ∫_a^bf(x).dx. For the first figure the function in inte ...

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Standard XII

Mathematics

Property 7

If area of sh...

Question

If area of shaded portion in figure A is $A_{1}$ and that of shaded portion in figure B is $A_{2}$ . Then,

A_{1} > A_{2}

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A_{1} < A_{2}

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A_{1} = A_{2}

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Solution

The correct option is C $A_{1} = A_{2}$
As we already know the area under the curve is given by the integral $\int_{a}^{b} f (x) . d x$ .
For the first figure the function in integral is f(x) = |x| and the concerned interval is from 0 to 1
$∴$ Area of the first figure = $\int_{0}^{1} x . d x = A_{1}$ .
Also, for the second figure the function in integral is f(x) = |t| and the concerned interval is from 0 to 1
Area of the second figure = $\int_{0}^{1} t . d t = A_{2}$ .
As seen in the video if we change just the variable of the integral without changing the nature of the function or the limits, the value of the integral doesn’t change. Here the function is modulus which is same for both the figures.
Hence, $\int_{0}^{1} | x | d x = \int_{0}^{1} | t | d t$
$A_{1} = A_{2}$

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If area of shaded portion in figure A is A1 and that of shaded portion in figure B is A2. Then,

If area of shaded portion in figure A is $A_{1}$ and that of shaded portion in figure B is $A_{2}$ . Then,