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 ;

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

${:\implies \quad \sf F(x)\bigg|_{0}^{6}=9}$

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

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

Also , we are given with ;

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

${:\implies \quad \sf F(x)\bigg|_{3}^{6}=5}$

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

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

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 ;

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

From the distributive of Integrals property we have ;

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

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

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

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 ;

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

${:\implies \quad \sf \dfrac{1}{2}(4)+3\times 7\quad \qquad \{\because (iii)\:\: and\:\: Given\}}$

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

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

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