Given that $\int_{1}^{3} f (x) d x = 3$ $int_1^3f(x)dx=3$ and $\int_{3}^{6} f (x) d x = 5$ $int_3^6f(x)dx=5$ , find: $\int_{1}^{3} [f (x) + x] d x$ $int_1^3[f(x)+x]dx$ ?

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Given that $\int_{1}^{3} f (x) d x = 3$ $int_1^3f(x)dx=3$ and $\int_{3}^{6} f (x) d x = 5$ $int_3^6f(x)dx=5$ , find: $\int_{1}^{3} [f (x) + x] d x$ $int_1^3[f(x)+x]dx$ ?

and in my book the solution is
$\int_{1}^{3} [f (x) + x] d x$ $int_1^3[f(x)+x]dx$
$= \int_{1}^{3} f (x) d x + \int_{1}^{3} x f (x)$ $=int_1^3f(x)dx+int_1^3xf(x)$
$= 3 + {[\frac{x^{2}}{2}]}^{3} = 7$ $=3+[x^2/2]^3 =7$
What is the confusion?

but when im solving it im getting 4 as an answer

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Answer 1 · 2018-02-03T20:45:18

See below.

Explanation:

We have:

$= \int_{1}^{3} f (x) d x + \int_{1}^{3} x d x$ $=int_1^3 f(x) dx+ int_1^3 x dx$

We know from the question that $\int_{1}^{3} f (x) d x = 3$ $int_1^3 f(x) dx = 3$ . Also, from the basic power rule of integration, we have $\int_{a}^{b} x d x = {[\frac{1}{2} x^{2}]}_{1}^{2}$ $int_a^b x dx = [1/2x^2]_1^3$ .

$= 3 + {[\frac{1}{2} x^{2}]}_{1}^{2}$ $=3 + [1/2x^2]_1^3$

$= 3 + \frac{1}{2} {(3)}^{2} - \frac{1}{2} {(1)}^{2}$ $=3 + 1/2(3)^2 - 1/2(1)^2$

$= 3 + \frac{9}{2} - \frac{1}{2}$ $=3 + 9/2 - 1/2$

$= 3 + \frac{8}{2}$ $=3 + 8/2$

$= 3 + 4$ $=3 + 4$

$= 7$ $=7$

Hopefully this helps!

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Given that $\int_{1}^{3} f (x) d x = 3$ $int_1^3f(x)dx=3$ and $\int_{3}^{6} f (x) d x = 5$ $int_3^6f(x)dx=5$ , find: $\int_{1}^{3} [f (x) + x] d x$ $int_1^3[f(x)+x]dx$ ?

and in my book the solution is
$\int_{1}^{3} [f (x) + x] d x$ $int_1^3[f(x)+x]dx$
$= \int_{1}^{3} f (x) d x + \int_{1}^{3} x f (x)$ $=int_1^3f(x)dx+int_1^3xf(x)$
$= 3 + {[\frac{x^{2}}{2}]}^{3} = 7$ $=3+[x^2/2]^3 =7$
What is the confusion?

but when im solving it im getting 4 as an answer

1 Answer

Explanation:

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Given that ∫31f(x)dx=3int_1^3f(x)dx=3 and ∫63f(x)dx=5int_3^6f(x)dx=5, find: ∫31[f(x)+x]dxint_1^3[f(x)+x]dx ?

and in my book the solution is ∫31[f(x)+x]dxint_1^3[f(x)+x]dx =∫31f(x)dx+∫31xf(x)=int_1^3f(x)dx+int_1^3xf(x) =3+[x22]3=7=3+[x^2/2]^3 =7 What is the confusion? but when im solving it im getting 4 as an answer

1 Answer

Explanation:

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Given that $\int_{1}^{3} f (x) d x = 3$ $int_1^3f(x)dx=3$ and $\int_{3}^{6} f (x) d x = 5$ $int_3^6f(x)dx=5$ , find: $\int_{1}^{3} [f (x) + x] d x$ $int_1^3[f(x)+x]dx$ ?

and in my book the solution is
$\int_{1}^{3} [f (x) + x] d x$ $int_1^3[f(x)+x]dx$
$= \int_{1}^{3} f (x) d x + \int_{1}^{3} x f (x)$ $=int_1^3f(x)dx+int_1^3xf(x)$
$= 3 + {[\frac{x^{2}}{2}]}^{3} = 7$ $=3+[x^2/2]^3 =7$
What is the confusion?

but when im solving it im getting 4 as an answer