Integration Flashcards
indefinite integration
“increase index by 1, then divide by new index”
ADD C
S ax^n dx =
a x^n+1/n+1 + c
A curve has gradient x3, and passes through the point (1, 4).
Find the equation of the curve
y = S x^3 dx = 0.25x^4 +c
4 = 0.25 (1)^4 + c
15/4 + c
y = 0.25x^4 + 15/4
Find A, if dA/dx = x^2 - 3x and A = 6 when x = 1
A = S x^2 - 3x dx = 1/3x^3 - 3/2x^2 + c
6 + 1/3(1)^3 - 3/2(1)2 + c
43/6 = c
A = 1/3x^3 = 3/2x^2 + 43/6
A curve has second derivative 3 - 6x
It passes through the point
(1,2), and its gradient at that point is -3. Find the equation of the curve.
-3 = c
d = 4.5
y = 3/2x^2 - x^3 - 3x + 9/2
FTC (fundamental theorem of calculus)
………….x
F (x) = S f(x) dx
a
is an antiderivative of f(x)
evaluating definite integrals without Riemann sums
b……….
S f(x) dx = F(b) - F(a)
a……….
where d/dx F(x) = f(x)
In definite integration we evaluate the integral at
two points, and take the difference
We call these two points the limits of integration
notate definite integration like this
limits are written at top and bottom, after the operator and square brackets
square brackets mean we’ve integrated but not applied limits yet
.5
S x + 1 dx =
.3
10
.
.
.1
S x^-3 + 4x^3 dx
.-1
0
.9
S 3/x^1/2 dx
.8
18 - 12 root2
area is always positive, but
definite integrals of regions below the axis are negative → we have to take the magnitude in such cases
how to find area under the curve
sketch curve (if not given)
find and marks limits
definite integration
split integrals
find magnitude of area below axis
