Question

Stuck on a calc problem: #30 points!

Answer 1

Let's first write the given we have :

• 
  `{\displaystyle \sf \int_(0)^(6)f(x)\: dx=9}`

• 
  `{\displaystyle \sf \int_(3)^(6)f(x)\: dx=5}`

• 
  `{\displaystyle \sf \int_(3)^(0)g(x)\: dx=-7}`

Now , let's assume that ;

`{:\implies \quad \displaystyle \sf \int f(x)\: dx=F(x)}`

Now , proceeding further ;

`{:\implies \quad \displaystyle \sf \int_(0)^(6)f(x)\: dx=9}`

`_(0)^(6)=9`

`{:\implies \quad \sf F(6)-F(0)=9}`

`{:\implies \quad \bf F(6)=F(0)+9\quad \qquad ---(i)}`

Also , we are given with ;

`{:\implies \quad \displaystyle \sf \int_(3)^(6)f(x)\: dx=5}`

`_(3)^(6)=5`

`{:\implies \quad \sf F(6)-F(3)=5}`

`{:\implies \quad \bf F(6)=F(3)+5\quad \qquad ---(ii)}`

Now from (i) & (ii)

`{:\implies \quad \sf F(3)+5=F(0)+9}`

`{:\implies \quad \sf F(3)-F(0)=9-5}`

`{:\implies \quad \bf F(3)-F(0)=4\quad \qquad ---(iii)}`

Now , let's go to what we have to find ;

`{:\implies \quad \displaystyle \sf \int_(0)^(3)\bigg\{(1)/(2)f(x)-3g(x)\bigg\}dx}`

From the distributive of Integrals property we have ;

`{:\implies \quad \displaystyle \sf \int_(0)^(3)(1)/(2)f(x)\: dx-\displaystyle \sf \int_(0)^(3)3g(x)\: dx}`

We knows that we can take out the Constant from the Integrand , So

`{:\implies \quad \displaystyle \sf (1)/(2)\int_(0)^(3)f(x)\: dx -3\int_(0)^(3)g(x)\: dx}`

Now , we knows a property of definite Integrals :

• 
  `{\boxed{\displaystyle \bf \int_(a)^(b)f(x)\: dx=-\displaystyle \bf \int_(b)^(a)f(x)\:dx}}`

Using this property and expanding the definite integral of f(x) we have ;

`{:\implies \quad \sf (1)/(2)\{F(3)-F(0)\}+3\displaystyle \sf \int_(3)^(0)g(x)\: dx}`

`{:\implies \quad \sf (1)/(2)(4)+3* 7\quad \qquad \{\because (iii)\:\: and\:\: Given\}}`

`{:\implies \quad \sf 2+21}`

`{:\implies \quad \bf 23}`

`{:\implies \quad \bf \therefore \quad \underline{\underline{\displaystyle \bf \int_(0)^(3)\bigg\{(1)/(2)f(x)-3g(x)\bigg\}dx=23}}}`