CUET UG 2025 — Mathematics Algebra

The area bounded by the curve $y = \log x$y=logx, $y = 0$y=0, and $x = e$x=e needs to be found.

The curve $y = \log x$y=logx crosses the x-axis when $\log x = 0$logx=0, which occurs at $x = 1$x=1 since $\log 1 = 0$log1=0.

The region extends from $x = 1$x=1 to $x = e$x=e.

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The area is given by:

Area $\displaystyle = \int_{1}^{e} \log x \, dx$=∫1e​logxdx

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Using the standard integration formula:

$$\int \log x \, dx = x \log x - x + C$$∫logxdx=xlogx−x+C

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Applying the limits:

Area $= [x \log x - x]_{1}^{e}$=[xlogx−x]1e​

At $x = e$x=e:

$$e \cdot \log e - e$$e⋅loge−e

$$= e \cdot 1 - e$$=e⋅1−e

$$= 0$$=0

At $x = 1$x=1:

$$1 \cdot \log 1 - 1$$1⋅log1−1

$$= 1 \cdot 0 - 1$$=1⋅0−1

$$= -1$$=−1

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Area $= 0 - (-1)$=0−(−1)

$$= 1$$=1

Therefore, the area bounded by the curve is $1$1 square unit.