Rules For Integration
Constant Multiple Rule
A function is an antiderivative of a constant multiple of of a function f if and only if it is k times an antiderivative of f. ∫ k f(x)dx k = ∫f(x)dx; where k is a constant
Question 1. \(\int 5 x^2 d x\)
Answer: \(\frac{5 x^3}{3}+C\)
Question 2. \(\int \frac{7}{x^2} d x\)
Answer:
⇒ \(\int 7 x^{-2} d x\)
⇒ \(-\frac{7 x^{-1}}{1}+C\)
⇒ \(\frac{-7}{x}+C\)
Question 3. \(\int \frac{t}{\sqrt{t}} d t\)
⇒ \(\int t^{1 / 2} d t=\frac{t^{3 / 2}}{3 / 2}+C\)
⇒ \(\frac{2}{3} t^{3 / 2}+C\)
Sum And Difference Rule
A function is an antiderivative of a sum or difference f ± g if and only if it is the sum or difference of an antiderivative of f or an antiderivative of g.
⇒ \(\int[f(x) \pm g(x)] d x=\int f(x) d x \pm \int g(x) d x\)
Question 4. Term–by–term integration. Evaluate : ∫(x2 – 2x + 5) dx.
Answer:
If we recognize that (x3 /3) – x2 + 5x is an antiderivative of x2 – 2x + 5, we can evaluate the integral as
If we do not recognize the antiderivative right away, we can generate it term by term with the sum and difference Rule:
⇒ \(\int\left(x^2-2 x+5\right) d x\)
⇒ \(\int x^2 d x-\int 2 x d x+\int 5 d x\)
⇒ \(\frac{x^3}{3}+C_1-x^2+C_2+5 x+C_3\)
This formula is more complicated than it needs to be. If we combine C1,C2 and C3 into a single constant C = C1+ C2+ C3, the formula simplifies to
⇒ \(\frac{x^3}{3}-x^2+5 x+C\)
and still gives all the antiderivatives there are. For this reason, we recommend that you go right to the final form even if you elect to integrate term by term. Write
⇒ \(\int\left(x^2-2 x+5\right) d x\)
⇒ \(\int x^2 d x-\int 2 x d x+\int 5 d x\)
⇒ \(\frac{x^3}{3}-x^2+5 x+C\)
Find the simplest antiderivative you can for each part and add the constant at the end.
Question 5. Find a body’s velocity from its acceleration and initial velocity. The acceleration of gravity near the surface of the earth is 9.8 m/sec2. This means that the velocity v of a body falling freely in a vacuum changes at the rate of \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 9.8 m/sec2. If the body is dropped from rest, what will its velocity be t seconds after it is released?
Answer:
In mathematical terms, we want to solve the initial value problem that consists of
The differential condition: \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 9.8
The initial condition: v = 0 when t = 0 ( abbreviated as v (0) = 0 )
We first solve the differential equation by integrating both sides concerning t:
⇒ \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 9.8 The differential equation
⇒ \(\int \frac{d v}{d t} d t=\int 9.8 d t\) Integrate with respect to t.
v + C1 = 9.8t + C2 Integrals evaluated
v = 9.8t + C. Constants combined as one
This last equation tells us that the body’s velocity t seconds into the fall is 9.8t + C m/sec. For value of C: What value? We find out from the initial condition:
v = 9.8t + C
0 = 9.8(0) + C v( 0) = 0
C = 0.
Conclusion: The body’s velocity t seconds into the fall is
v = 9.8t + 0 = 9.8t m/sec.
- The indefinite integral F(x) + C of the function f(x) gives the general solution y = F(x) + C of the differential equation dy/dx = f(x).
- The general solution gives all the solutions of the equation (there are infinitely many, one for each value of C).
- We solve the differential equation by finding its general solution We then solve the initial value problem by finding the particular solution that satisfies the initial condition y(xo) = yo( y has the value yo when x = xo.).
Definite Integration Or Integration With Limits
⇒ \(\int_a^b f(x) d x=[g(x)]_a^b=g(b)-g(a)\)
where g(x) is the antiderivative of f(x) i.e. g´(x) = f(x)
Question 6. \(\int_{-1}^4 3 d x\)
Answer:
⇒ \(3 \int_{-1}^4 d x\)
⇒ \(3[x]_{-1}^4\)
⇒ \(3[4-(-1)]\)
= (3)(5)=15
⇒ \(\int_0^{\pi / 2} \sin x d x\)
⇒ \([-\cos x]_0^{\pi / 2}\)
⇒ \(-\cos \left(\frac{\pi}{2}\right)+\cos (0)\)
= –0 + 1 = 1
Application Of Definite Integral: Calculation Of Area Of A Curve
From the graph shown in the figure if we divide the whole area into infinitely small strips of dx width. We take a strip at x position of dx width.
A small area of this strip dA = f(x) dx
So, the total area between the curve and x-axis = sum of area of all strips = \(\int_a^b f(x) d x\)
Let f(x) ≥ 0 be continuous on [a,b]. The area of the region between the graph of f and the x-axis is
A = \(\int_a^b f(x) d x\)
Question 7. Find the area under the curve of y = x from x = 0 to x = a
Answer:
\(\int_0^a y d x=\left.\frac{x^2}{2}\right|_0 ^a=\frac{a^2}{2}\)