How to find integration??

Question

How to find integration??

Given that $\int_{0}^{4} h (x) d x = 5$ $int_0^4 h(x) dx = 5$ , evaluate:

a) $\int_{0}^{4} 2 h (x) d x$ $int_0^4 2h(x) dx$

b) $\int_{0}^{4} h (x) - x d x$ $int_0^4 h(x)-x dx$

c) $\int_{0}^{4} h (x) + 1 d x$ $int_0^4 h(x) +1 dx$

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Answer 1 · 2018-04-15T09:33:09

a) 10
b) -3
c) 9

Explanation:

a) Remember that a constant can be written outside the integral. Therefore,:
$\int_{0}^{4} 2 h (x) d x = 2 \cdot \int_{0}^{4} h (x) d x = 2 \cdot 5 = 10$ $int_0^4 2h(x) dx=2*int_0^4 h(x) dx=2*5=10$

b) Remember that $\int (f (x) + g (x)) = \int (f (x)) + \int (g (x))$ $int (f(x)+g(x))=int(f(x))+int(g(x))$

Therefore:
$\int_{0}^{4} h (x) - x d x = \int_{0}^{4} h (x) d x - \int_{0}^{4} x d x$ $int_0^4 h(x)-x dx=int_0^4 h(x)dx-int_0^4x dx$
$\int_{0}^{4} x d x = {[\frac{1}{2} x^{2}]}_{0}^{2} = \frac{16}{2} - \frac{1}{2} \cdot 0 = 8$ $int_0^4x dx=[1/2x^2]_0^4=16/2-1/2*0=8$
$\int_{0}^{4} h (x) d x - \int_{0}^{4} x d x = 5 - 8 = - 3$ $int_0^4 h(x)dx-int_0^4x dx=5-8=-3$

c) $\int_{0}^{4} h (x) + 1 d x = \int_{0}^{4} h (x) d x + \int_{0}^{4} 1 d x$ $int_0^4 h(x) +1 dx=int_0^4 h(x)dx +int_0^4 1 dx$
$\int_{0}^{4} 1 d x = {[x]}_{0}^{4} = 4 - 0 = 4$ $int_0^4 1 dx=[x]_0^4=4-0=4$
$\int_{0}^{4} h (x) d x + \int_{0}^{4} 1 d x = 5 + 4 = 9$ $int_0^4 h(x)dx +int_0^4 1 dx=5+4=9$

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How to find integration??

Given that $\int_{0}^{4} h (x) d x = 5$ $int_0^4 h(x) dx = 5$ , evaluate:

a) $\int_{0}^{4} 2 h (x) d x$ $int_0^4 2h(x) dx$

b) $\int_{0}^{4} h (x) - x d x$ $int_0^4 h(x)-x dx$

c) $\int_{0}^{4} h (x) + 1 d x$ $int_0^4 h(x) +1 dx$

1 Answer

Explanation:

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How to find integration??

Given that ∫40h(x)dx=5int_0^4 h(x) dx = 5, evaluate: a) ∫402h(x)dxint_0^4 2h(x) dx b) ∫40h(x)−xdxint_0^4 h(x)-x dx c) ∫40h(x)+1dxint_0^4 h(x) +1 dx

1 Answer

Explanation:

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Impact of this question

Given that $\int_{0}^{4} h (x) d x = 5$ $int_0^4 h(x) dx = 5$ , evaluate:

a) $\int_{0}^{4} 2 h (x) d x$ $int_0^4 2h(x) dx$

b) $\int_{0}^{4} h (x) - x d x$ $int_0^4 h(x)-x dx$

c) $\int_{0}^{4} h (x) + 1 d x$ $int_0^4 h(x) +1 dx$