LEIBNITZ’S RULE
z If f(x) is continuous and u(x), v(x) are differentiable
functions in the interval [a, b], then,
d
dx ftdt fvx
d
dxvx f uxd
ux dxuxvx
()( ) { ( )} { ( )} { ( )} { ( )}()
∫ = −z If the function φ(x) and Ψ(x) are defined on [a,b]
and differentiable at a point x ∈(a, b) and f(x, t) is
continuous, then,
d
dx
fxtdt d
dxfxtdt dx
x dxx
xx
() (,) (,) ()()
()()
φφΨΨ Ψ
∫∫⎡
⎣⎢⎤
⎦⎥=+⎧
⎨
⎩⎫⎫
⎬
⎭×fx x dx
dx(, ())Ψ−φ()fx x(, ())φAPPLICATION OF INTEGRALS
(i) The area bounded
by the curve y = f(x),
the x-axis and the lines
x = a and x = b is Afxdx
ab
=∫|()|(ii) The area bounded
by the curves y = f(x),
y = g(x) and the lines
x = a and x = b is
Afxgxdx
ab
=∫|() ()|−(iii) The area bounded
by the curve x = f(y),
the y-axis and the lines
y = c and y = d is
Afydy
cd
=∫|()|(iv) The area bounded
by the curves x = f(y)
and x = g(y) and the
lines y = c and y = d is
Afygydy
cd
=∫|() ()|−PROBLEMS
Single Correct Answer Type- sin
sin( )
x
xdx
−
∫ =
α
(a) xcosα – sinα logsin(x – α) + c
(b) xcosα + sinα logsin(x – α) + c
(c) xsinα – sinα logsin(x – α) + c
(d) None of these- cos
cos
21
21x
x−dx
+
∫ =
(a) tanx – x + c (b) x + tanx + c
(c) x – tanx + c (d) –x – cotx + c- ()
()
x
xx+ dx
∫ +1
12
2 is equal to
(a) logex + c (b) logex + 2tan–1x + c
(c) loge
x(^1) c
(^2) + 1
⎛
⎝⎜
⎞
⎠⎟
- (d) loge{x(x^2 + 1)} + c
- xe
xe
dxex
ex−−+
+
∫ =11(a) log(xe + ex) + c (b) elog(xe + ex) + c
(c)^1
elog(xe cex++) (d) None of these- sin
sin cos
2
44x
xxdx
+
∫ =
(a) cot–1(tan^2 x) + c (b) tan–1(tan^2 x) + c
(c) cot–1(cot^2 x) + c (d) tan–1(cot^2 x) + c- sin
sin
2
222x
ab xdx
+
∫ =(a)^12222
blog(ab++sin xc)(b)^1222
blog(ab++sin xc)
(c) log(a^2 + b^2 sin^2 x) + c
(d) b^2 log(a^2 + b^2 sin^2 x) + c- cos
(cos sin )
2
2x
xxdx
+
∫ =(a) log cosxxc++sin (b) log(cosx – sinx) + c
(c) log(cosx + sinx) + c (d) −
+(^1) +
cosxxsin
c
8.
1
1 −^2
∫ =
e
dx
x
(a) xec−log[ 11 + −^2 x]+
(b) xec++log[^11 −^2 x]+
(c) log[ 11 + −−exc^2 x] + (d) None of these
- sin
sin sin
2
53x
xx
∫ dx=