Question

find first quadrant area bounded by the curves using integration :
y = arctanx, y=π/4 and x=0.​

Answer 1

we have to find the area bounded by the curves using integration : y = tan¯¹x , y = π/4 and x = 0.

Solution : as we have to find area enclosed by y = tan¯¹x , y = π/4 and x = 0, shaded region shown in figure is the required area formed by given curves.

so upper limit , y₂ = π/4

lower limit, y₁ = 0

now area bounded by curves = $\int\limits^{\pi/4}_0{tanx}\,dx$

= $[ln|secx|]^{\pi/4}_0$

= ln(√2) - ln(1)

= ln(√2)

≈ 0.346

Therefore the area bounded by curve is 0.346 sq unit

Answer 2

The area bounded by the curve y = arctanx from y = π/4 and x=0. is ㏑2/2 or 0.347

Step-by-step explanation:

Given function

$y=\tan^{-1}x$

When $y=\frac{\pi}{4}$, $x=1$

The area bounded will be

$\int_0^{1}\tan^{-1}xdx$

Lets first find the indefinite integral $\int \tan^{-1}xdx$

$\int \tan^{-1}xdx$

$=\int \tan^{-1}x.1dx$

$=\int \tan^{-1}x.1dx$

$=\tan^{-1}x\int 1.dx-\int [(\frac{d}{dx}(\tan^{-1}x)\int 1.dx]dx$

$=x\tan^{-1}x-\int [\frac{1}{1+x^2}\times x]dx$

$=x\tan^{-1}x-\int \frac{x}{1+x^2}dx$

$=x\tan^{-1}x-\frac{1}{2}\int \frac{2x}{1+x^2}dx$

$=x\tan^{-1}x-\frac{1}{2}\ln (1+x^2)+C$    where C is a constant

Now

$\int_0^{1}\tan^{-1}xdx$

$=\Bigr |x\tan^{-1}x-\frac{1}{2}\ln (1+x^2)+C\Bigr |_0^{1}$

$=\frac{\pi}{4}\times 1-\frac{\ln(1+1^2)}{2}-0$

$=\frac{\pi}{4}-\frac{1}{2}\ln 2$

But then this is the area bounded by the lines x = 0 and x = 1

While we need to find out the area bounded by x = 0 and y = π/4

So we need to subtract this area from  the area of the rectangle whose sides are π/4  and 1

Thus our required area bounded by the given curve will be

$\frac{\pi}{4}.1-(\frac{\pi}{4}-\frac{1}{2}\ln 2)$

$=\frac{1}{2}\ln 2$

$=0.347$ sq units

Hope this answer is helpful.

Know More:

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