Please only answer the red outlined boxes Using Properties of Definite Integrals.2Given| f(x) dx = 0 andf(x) dx = 4-2evaluate(a)f(x)dx =-2(b)f(x) dx -f(x) dx =(c)3f(x) dx = 0%3D-2(d)3f(x) dx = 12%3D

we know that ∫jkfxdx=∫jlfxdx+∫lkfxdx therefore ∫-22fxdx=∫-20fxdx+∫02fxdx substitute the values…

Answered: ng Properties of Definite Integrals. 2…

Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Please only answer the red outlined boxes

Using Properties of Definite Integrals.
2
Given
| f(x) dx = 0 and
f(x) dx = 4
-2
evaluate
(a)
f(x)
dx =
-2
(b)
f(x) dx -
f(x) dx =
(c)
3f(x) dx = 0
%3D
-2
(d)
3f(x) dx = 12
%3D

Expert Solution

Step 1

we know that $\int_{j}^{k} f (x) d x = \int_{j}^{l} f (x) d x + \int_{l}^{k} f (x) d x$

therefore

$\int_{- 2}^{2} f (x) d x = \int_{- 2}^{0} f (x) d x + \int_{0}^{2} f (x) d x$

substitute the values

$\begin{array}{rcl} \int_{- 2}^{2} f (x) d x & = & \int_{- 2}^{0} f (x) d x + \int_{0}^{2} f (x) d x \\ 0 & = & \int_{- 2}^{0} f (x) d x + 4 \\ 0 - 4 & = & \int_{- 2}^{0} f (x) d x \\ - 4 & = & \int_{- 2}^{0} f (x) d x \end{array}$

hence $\begin{array}{rcl} \int_{- 2}^{0} f (x) d x & = & - 4 \end{array}$

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ng Properties of Definite Integrals. 2 .2 en f(x) dx = 0 and f(x) dx = 4