Application of Integrals Test - 30

For $$x=1,y=b.{5}^{x}+4=5b+4$$ and 
$$\dfrac{dy}{dx}=b.{5}^{x}\log{5}$$
$$\Rightarrow 5b\log{5}=40\log{5}$$
$$\Rightarrow b=8$$
The two curves intersect at points where
$$8.{5}^{x}+4={25}^{x}+16$$
$$\Rightarrow {5}^{2x}-8.{5}^{x}+12=0$$
On factorizing we get
$$x=\log_{5}{2},x=\log_{5}{6}$$
Hence the area of the given region
$$=\int_{\log_{5}{2}}^{\log_{5}{6}}{\left[8.{5}^{x}-{25}^{x}-12\right]dx}$$
$$=\left[\dfrac{8.{5}^{x}}{\log_{e}{5}}-12x-\dfrac{{25}^{x}}{\log_{e}{25}}\right]_{\log_{5}{2}}^{\log_{5}{6}}$$
$$=\dfrac{-{5}^{\log_{5}{36}}}{\log_{e}{5}}+\dfrac{8.{5}^{x}}{\log_{e}{5}}-12\left(\log_{5}{6}-\log_{5}{2}\right)+\dfrac{{5}^{\log_{5}{4}}}{\log_{e}{25}}-\dfrac{16}{\log_{e}{5}}$$
On simplification, we get
$$=\dfrac{16}{\log_{e}{5}}-12\log_{e}{3}$$
$$=4\log_{5}{{e}^{4}}-4\log_{5}{27}$$
$$=4\log_{5}{\left(\dfrac{{e}^{4}}{27}\right)}$$

Given that,

The curve and the coordinate of any point on its equal to the cube of

that ordinates.

Let (h, k) be point on it is equal to the cube of that ordinate.

Given the area of the region bounded by the co-ordinate axis in above figure.

We know that the area of rectangle is

$$Area=length\times breath$$

But Given that ,

Ordinate of point $$=k$$

$$h\times k={{k}^{3}}\,$$                  (given function )

$$h={{k}^{2}}$$

Hence it represent the parabola.

$${x^2} + {y^2} = 16{a^2} - \left( 1 \right)$$

$$\int_{0}^{t}{\left(f\left(x\right)-\left({x}^{4}-4{x}^{2}\right)\right)dx}=k\int_{0}^{t}{\left(2{x}^{2}-{x}^{3}-f\left(x\right)\right)dx}$$
Differentiating w.r.t $$t$$ we get
$$f\left(t\right)-{t}^{4}+4{t}^{2}=k\left(2{t}^{2}-{t}^{3}-f\left(t\right)\right)$$
$$\Rightarrow \left(1+k\right)f\left(t\right)={t}^{4}-4{t}^{2}+k\left(2{t}^{2}-{t}^{3}\right)$$
$$\therefore f\left(t\right)=\dfrac{{t}^{4}-4{t}^{2}+k\left(2{t}^{2}-{t}^{3}\right)}{k+1}$$
$$\int_{-1}^{1}{f\left(x\right)dx}=\int_{-1}^{1}{\dfrac{{x}^{4}-4{x}^{2}+k\left(2{x}^{2}-{x}^{3}\right)}{k+1}dx}$$
$$=\int_{-1}^{1}{\dfrac{{x}^{4}-k{x}^{3}+\left(2k-4\right){x}^{2}}{k+1}dx}$$
We know that $$\int_{-a}^{a}{f\left(x\right)dx}=0$$ if $$f\left(x\right)=-f\left(x\right)$$
Thus,$$\int_{-1}^{1}{{x}^{3}dx}=0$$.Thus the above integral becomes,
$$\int_{-1}^{1}{\dfrac{{x}^{4}+\left(2k-4\right){x}^{2}}{k+1}dx}$$
Using the concept $$\int_{-a}^{a}{f\left(x\right)dx}=2\int_{0}^{a}{f\left(x\right)dx}$$ we have
$$=\dfrac{2}{k+1}\left[\dfrac{{x}^{5}}{5}\right]_{0}^{1}+\dfrac{2k-4}{k+1}\times 2\left[\dfrac{{x}^{3}}{3}\right]_{0}^{1}$$
$$=\dfrac{2}{k+1}\times \dfrac{1}{5}+\dfrac{4k-8}{k+1}\dfrac{1}{3}$$
$$=\dfrac{6+20k-40}{15\left(k+1\right)}$$
$$=\dfrac{20k-34}{15\left(k+1\right)}=\dfrac{2\left(10k-17\right)}{15\left(k+1\right)}$$

$$A=$$area$$(OAB)-$$ area of$$\triangle{OAB}\\=\cfrac{1}{4}(\pi. 2^2)-\cfrac{1}{2}\times 2\times 2\\=\pi-2$$

$$A=\int _{ 0 }^{ \pi }x+\sin{x}-xdx$$