Question #270706

Q.1: What are the applications of Calculus in engineering?

Q.1: Define differentiation and integration with example. What are the differences between them?

Q.3: Integrate the following functions with respect to x:

       Sin3x, x^6  , xy, e^5x  , 10 .

Q.4: Describe geometrical meaning of indefinite integral. Write down

       some properties of indefinite integral.


Expert's answer

 Q.1

i) Computing the surface area of complex objects to determine frictional forces

ii) Designing a pump according to flow rate and head

iii) Calculating the power provided by a battery system

Q.2

Differentiation is a process of finding the instantaneous rate of change in function based on one of its variables. For Example, differentiate the function y=x2y=x^2 .Then our solution is dydx=2x{dy\over dx}=2x

Integration is a method to find definite and indefinite integrals. The integration of a function f(x) is given by F(x)

For example, integrate the function 2x2x .Then our solution is 2xdx=2x1+11+1=x2\int 2xdx={2x^{1+1}\over {1+1}}=x^2

The differences between differentiation and Integration is that Integration is the reverse of differentiation

Q.3:

i) sin3xdx\int sin3xdx

Let u=3x    dudx=3    dx=13duu=3x \implies {du\over dx}=3 \implies dx={1\over 3}du

Now, Substitute

=13sin(u)du={1\over 3}\int sin(u)du

=13[cos(u)]={1\over 3}[-cos(u)]

=13[cos(3x)]={1\over 3}[-cos(3x)]

sin3xdx=cos(3x)3+C\therefore \int sin3xdx= {-cos(3x)\over 3}+C

ii) x6dx=x6+16+1=17x7+C\int x^6dx={x^{6+1}\over 6+1}={1\over 7}x^7+C

iii) xydx=yxdx=yx1+11+1=12yx2+C\int xydx =y\int xdx=y{x^{1+1}\over 1+1}={1\over 2}yx^2+C

iv) e5xdx\int e^{5x}dx

Let u=5x    dudx=5    dx=du5u=5x \implies {du\over dx}=5 \implies dx= {du\over 5}

Substitute

eudu5=eu5+C=e5x5+C\int e^{u}{du\over 5}={e^u\over 5}+C={e^{5x}\over 5}+C

v) 10dx=10dx=10x+C\int 10dx=10\int dx=10x+C

Q.4:

f(x)dx=F(x)+C\int f(x) dx =F(x)+C represents a family of curves and different values of C corresponds to different members of this family.

Properties of indefinite integral

1) cf(x)dx=cf(x)dx\int cf(x)dx=c\int f(x)dx

From the indefinite integral, we can take out the multiplicative constants.

2) f(x)dx=f(x)dx\int -f(x)dx=\int f(x)dx

Due to the negative function, the indefinite integral is also negative.

3) [f(x)±g(x)]dx=f(x)dx±g(x)dx\int [f(x) \pm g(x)]dx=\int f(x)dx\pm \int g(x)dx

It shows the sum as well as the difference of the integral of the functions as the sum or the difference of their individual integral.








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